tutorial class iii and iv 17-29/09/2012 1. first order...

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.Tutorial Class III and IV 17-29/09/2012..

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...1 First Order Partial Derivatives;

...2 Higher Order Partial Derivatives;

...3 Linear Approximation

...4 Chain Rule of Partial Differentiation

...5 Implicit Function and Its Partial Derivatives;

...6 Maximum and Minimum;

...7 Critical Points, and Gradient of Scalar Function;

...8 Directional Derivative;

...9 Application of Gradient;

...10 Weekly Quiz.

Prepared by Dr. I.T. Leong for Tutorial 3-4 of Matb 210

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Example. Show that the function u(x, y) = ex sin y andv(x, y) = ex cos y are solutions of Laplace’s equation uxx + uyy = 0.

Solution. (i) First we have ux = ex sin y, and uxx = ex sin y.

Similarly, uy = −ex cos y, and uyy = −ex sin y.Hence uxx + uyy = ex sin y − ex sin y = 0........Work it out (in 1 minute only!)

(ii) vx = ex cos y and vxx = ex cos y.Similarly, vy = −ex sin y, and uyy = −ex cos y.Hence, vxx + vyy = ex cos y − ex cos y = 0.


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Solution. (i) First we have ux = ex sin y, and uxx = ex sin y.Similarly, uy = −ex cos y, and uyy = −ex sin y.

Hence uxx + uyy = ex sin y − ex sin y = 0........Work it out (in 1 minute only!)



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Solution. (i) First we have ux = ex sin y, and uxx = ex sin y.Similarly, uy = −ex cos y, and uyy = −ex sin y.Hence uxx + uyy = ex sin y − ex sin y = 0.

.

......Work it out (in 1 minute only!)



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Solution. (i) First we have ux = ex sin y, and uxx = ex sin y.Similarly, uy = −ex cos y, and uyy = −ex sin y.Hence uxx + uyy = ex sin y − ex sin y = 0........Work it out (in 1 minute only!)



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(ii) vx = ex cos y and vxx = ex cos y.

Similarly, vy = −ex sin y, and uyy = −ex cos y.Hence, vxx + vyy = ex cos y − ex cos y = 0.


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(ii) vx = ex cos y and vxx = ex cos y.Similarly, vy = −ex sin y, and uyy = −ex cos y.

Hence, vxx + vyy = ex cos y − ex cos y = 0.


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Example. Let u(x, y, z) = ln(x3 + y3 + z3 − 3xyz), prove thatux + uy + uz =

3x+y+z on the domain where all these functions are

well-defined.

Solution.

ux =1

x3 + y3 + z3 − 3xyz· ∂

∂x(x3 + y3 + z3 − 3xyz)

=3x2 − 3yz

x3 + y3 + z3 − 3xyz,

uy =3y2 − 3xz

x3 + y3 + z3 − 3xyz, uz =

3z2 − 3xyx3 + y3 + z3 − 3xyz

.

ux + uy + uz =3(x2 + y2 + z2 − xy − yz − xz)

x3 + y3 + z3 − 3xyz=

3x + y + z

.

The last equality follows from the algebraic identity:x3 + y3 + z3 − 3xyz = (x + y + z)(x2 + y2 + z2 − xy − yz − xz).


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well-defined.

Solution.

ux =1

x3 + y3 + z3 − 3xyz· ∂

∂x(x3 + y3 + z3 − 3xyz)

=3x2 − 3yz

x3 + y3 + z3 − 3xyz,

uy =3y2 − 3xz

x3 + y3 + z3 − 3xyz, uz =

3z2 − 3xyx3 + y3 + z3 − 3xyz

.


x3 + y3 + z3 − 3xyz

=3

x + y + z.



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well-defined.

Solution.

ux =1

x3 + y3 + z3 − 3xyz· ∂

∂x(x3 + y3 + z3 − 3xyz)

=3x2 − 3yz

x3 + y3 + z3 − 3xyz,

uy =3y2 − 3xz

x3 + y3 + z3 − 3xyz, uz =

3z2 − 3xyx3 + y3 + z3 − 3xyz

.


x3 + y3 + z3 − 3xyz=

3x + y + z

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Example. If z(x, y) = y + f (x2 − y2) where f is a differentiable functionof one variable on R, prove that y ∂z

∂x + x ∂z∂y = x.

Solution. This follows from chain rule that∂z∂x

=∂

∂x( y + f (x2 − y2) ) = f ′(x2 − y2)

∂

∂x(x2 − y2) = 2xf ′(x2 − y2),

and similarly,∂z∂y

=∂

∂y( y+ f (x2 − y2) ) = 1+ f ′(x2 − y2)

∂

∂y(x2 − y2) = 1− 2yf ′(x2 − y2).

Then we have

y∂z∂x

+ x∂z∂y

= y · 2xf ′(x2 − y2) + x · (1 − 2yf ′(x2 − y2)) = x.


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Remarks of determining the max/min values of f ....1 In general, we need some conditions to ensure that the

extremum values of f exist before we start to find them, such as,the region D discussed is close and bounded.For example, in R2, the region D is bounded by a simple closedcurve; in R3, this is more difficult as it is hard to have an analog ofsimple closed surface. In general, it is easy to check that a regionis bounded, but to check that a region is closed, one needs tocheck its compliment is open, which is technical in nature.

...2 Similar to the one variable case, extremum values can occur atthe boundary points or interior critical points.


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Example. Find abs. max. and min. values of f (x, y) = x2 − 2xy + 2yon the rectangle R = { (x, y) | 0 ≤ x ≤ 3, 0 ≤ y ≤ 2}.

Solution. Since f is a polynomial, it is continuous on the closed,bounded rectangle R, and hence f attains both max. and min. on D.For interior critical point of f : (0, 0) = ∇f (x, y) = (2x − 2y,−2x + 2),i.e. (x, y) = (1, 1). And f (1, 1) = 1.

For the boundary point, let A(0, 0), B(3, 0), C(3, 2) and D(0, 2).(i) On AB, f (x, 0) = x2 for 0 ≤ x ≤ 3, and

0 = f (0, 0) ≤ f (x, 0) ≤ f (3, 0) = 9.(ii) On BC, f (3, y) = 9 − 6y + 2y = 9 − 4y for 0 ≤ y ≤ 2, and

1 = f (3, 2) ≤ f (3, y) ≤ f (3, 0) = 9.(iii) On CD, we have f (x, 2) = x2 − 4x + 4 = (x − 2)2 for 0 ≤ x ≤ 3,

and 0 = f (2, 2) ≤ f (x, 2) ≤ f (3, 0) = 9.(iv) On DA, we have f (0, y) = 2y for 0 ≤ y ≤ 2, and

0 = f (0, 0) ≤ f (3, y) ≤ f (0, 2) = 4.The abs. max. of f on R is the max. value determined by the interiorcritical points or the ones in the boundary of R, which is f (3, 0) = 9.Similarly, abs. min. value of f on R is given by f (0, 0) = 0.


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Solution. Since f is a polynomial, it is continuous on the closed,bounded rectangle R, and hence f attains both max. and min. on D.For interior critical point of f : (0, 0) = ∇f (x, y) = (2x − 2y,−2x + 2),i.e. (x, y) = (1, 1). And f (1, 1) = 1.For the boundary point, let A(0, 0), B(3, 0), C(3, 2) and D(0, 2).(i) On AB, f (x, 0) = x2 for 0 ≤ x ≤ 3, and

0 = f (0, 0) ≤ f (x, 0) ≤ f (3, 0) = 9.

(ii) On BC, f (3, y) = 9 − 6y + 2y = 9 − 4y for 0 ≤ y ≤ 2, and1 = f (3, 2) ≤ f (3, y) ≤ f (3, 0) = 9.

(iii) On CD, we have f (x, 2) = x2 − 4x + 4 = (x − 2)2 for 0 ≤ x ≤ 3,and 0 = f (2, 2) ≤ f (x, 2) ≤ f (3, 0) = 9.

(iv) On DA, we have f (0, y) = 2y for 0 ≤ y ≤ 2, and0 = f (0, 0) ≤ f (3, y) ≤ f (0, 2) = 4.

The abs. max. of f on R is the max. value determined by the interiorcritical points or the ones in the boundary of R, which is f (3, 0) = 9.Similarly, abs. min. value of f on R is given by f (0, 0) = 0.


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1 = f (3, 2) ≤ f (3, y) ≤ f (3, 0) = 9.

(iii) On CD, we have f (x, 2) = x2 − 4x + 4 = (x − 2)2 for 0 ≤ x ≤ 3,and 0 = f (2, 2) ≤ f (x, 2) ≤ f (3, 0) = 9.




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and 0 = f (2, 2) ≤ f (x, 2) ≤ f (3, 0) = 9.




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and 0 = f (2, 2) ≤ f (x, 2) ≤ f (3, 0) = 9.(iv) On DA, we have f (0, y) = 2y for 0 ≤ y ≤ 2, and

0 = f (0, 0) ≤ f (3, y) ≤ f (0, 2) = 4.The abs. max. of f on R is the max. value determined by the interiorcritical points or the ones in the boundary of R, which is f (3, 0) = 9.Similarly, abs. min. value of f on R is given by f (0, 0) = 0.


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......Example. The absolute maximum of f (x, y) = 2xy on x2 + y2

4 ≤ 18 isA. 22 B. 36 C. 54 D. 18 E. 40

Solution. First we want to determine the interior critical point of f , so

(0, 0) = ∇f (x, y) = (2y, 2x) if and only if (x, y) = (0, 0).

Note that f (0, 0) = 0.

Next to check the points on the boundary of the regionR : x2 + y2

4 ≤ 18, which can be parameterized by

r(t) = (x(t), y(t)) = (3√

2 cos t, 6√

2 sin t) where 0 ≤ t ≤ 2π.

It follows that

f (r(t)) = 2(3√

2 cos t)(6√

2 sin t) = 36(2 sin t cos t) = 36 sin(2t),

which has a maximum f (r(π/4)) = 36 ≥ f (0, 0) = 0 when t = π/4.


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......Example. The absolute maximum of f (x, y) = 2xy on x2 + y2

4 ≤ 18 isA. 22 B. 36 C. 54 D. 18 E. 40

Solution. First we want to determine the interior critical point of f , so

(0, 0) = ∇f (x, y) = (2y, 2x) if and only if (x, y) = (0, 0).

Note that f (0, 0) = 0.Next to check the points on the boundary of the regionR : x2 + y2

4 ≤ 18, which can be parameterized by

r(t) = (x(t), y(t)) = (3√

2 cos t, 6√

2 sin t) where 0 ≤ t ≤ 2π.

It follows that

f (r(t)) = 2(3√

2 cos t)(6√

2 sin t) = 36(2 sin t cos t) = 36 sin(2t),

which has a maximum f (r(π/4)) = 36 ≥ f (0, 0) = 0 when t = π/4.


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Example. Prove that the function f (x, y) = (1 + exp y) cos x − y exp yhas infinitely many local maxima, but no local minima at all in itsdomain.

Solution. ∇f (x, y) = ( − (1 + ey) sin x, ey cos x − yey − ey )

= ( − (1 + ey) sin x, ey(cos x − 1 − y) ).

As ey > 0, ∇f (x, y) = (0, 0) if and only if (sin x, cos x − 1 − y) = (0, 0),i.e. (x, y) = (nπ, (−1)n − 1) for some integer n.A = fxx(x, y) = −(1 + ey) cos x, C = fyy(x, y) = ey(cos x − 2 − y) )and B = fxy(x, y) = −ey cos x.At (x, y) = (2nπ, 0), we have A = −2, B = 0, C = −1 andAC − B2 = 2 > 0, hence f (2nπ, 0) = 2 are all local maxima.

Moreover, at the points (x, y) = ( (2n + 1)π, 0), we haveA = 1 + exp(−2), B = 0, C = − exp(−2) andAC − B2 = − exp(−2)− exp(−4) > 0, hence f ( (2n + 1)π, 0) are notlocal minima at all.Remark. The function g(|M|) = f (π/2, |M|) = −|M| exp(|M|)approaches to −∞ as |M| → +∞, hence f (x, y) does not have globalminimum.


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= ( − (1 + ey) sin x, ey(cos x − 1 − y) ).As ey > 0, ∇f (x, y) = (0, 0) if and only if (sin x, cos x − 1 − y) = (0, 0),i.e.

(x, y) = (nπ, (−1)n − 1) for some integer n.A = fxx(x, y) = −(1 + ey) cos x, C = fyy(x, y) = ey(cos x − 2 − y) )and B = fxy(x, y) = −ey cos x.At (x, y) = (2nπ, 0), we have A = −2, B = 0, C = −1 andAC − B2 = 2 > 0, hence f (2nπ, 0) = 2 are all local maxima.



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= ( − (1 + ey) sin x, ey(cos x − 1 − y) ).As ey > 0, ∇f (x, y) = (0, 0) if and only if (sin x, cos x − 1 − y) = (0, 0),i.e. (x, y) = (nπ, (−1)n − 1) for some integer n.

A = fxx(x, y) = −(1 + ey) cos x, C = fyy(x, y) = ey(cos x − 2 − y) )and B = fxy(x, y) = −ey cos x.At (x, y) = (2nπ, 0), we have A = −2, B = 0, C = −1 andAC − B2 = 2 > 0, hence f (2nπ, 0) = 2 are all local maxima.



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= ( − (1 + ey) sin x, ey(cos x − 1 − y) ).As ey > 0, ∇f (x, y) = (0, 0) if and only if (sin x, cos x − 1 − y) = (0, 0),i.e. (x, y) = (nπ, (−1)n − 1) for some integer n.A = fxx(x, y) = −(1 + ey) cos x, C = fyy(x, y) = ey(cos x − 2 − y) )and B = fxy(x, y) = −ey cos x.

At (x, y) = (2nπ, 0), we have A = −2, B = 0, C = −1 andAC − B2 = 2 > 0, hence f (2nπ, 0) = 2 are all local maxima.



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= ( − (1 + ey) sin x, ey(cos x − 1 − y) ).As ey > 0, ∇f (x, y) = (0, 0) if and only if (sin x, cos x − 1 − y) = (0, 0),i.e. (x, y) = (nπ, (−1)n − 1) for some integer n.A = fxx(x, y) = −(1 + ey) cos x, C = fyy(x, y) = ey(cos x − 2 − y) )and B = fxy(x, y) = −ey cos x.At (x, y) = (2nπ, 0), we have A = −2, B = 0, C = −1 andAC − B2 = 2 > 0, hence f (2nπ, 0) = 2 are all local maxima.



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Moreover, at the points (x, y) = ( (2n + 1)π, 0), we haveA = 1 + exp(−2), B = 0, C = − exp(−2) andAC − B2 = − exp(−2)− exp(−4) > 0, hence f ( (2n + 1)π, 0) are notlocal minima at all.

Remark. The function g(|M|) = f (π/2, |M|) = −|M| exp(|M|)approaches to −∞ as |M| → +∞, hence f (x, y) does not have globalminimum.


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Definition. For any differentiable f and a point P(a, b) in its domain, let

L(x, y) = f (a, b) + fx(a, b)(x − a) + fy(a, b)(y − b)

= f (a, b) +∇f (a, b) · (x − a, y − b)

be the linear approximation of f at the point P(a, b).

Remark. For any fixed P(a, b), the function L(x, y) associated to afunction f at P(a, b) has a graph z = L(x, y) which is the tangent plane..

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Example. Approximate the number√(3.2)2 + (3.9)2 using the linear

approximation to the function f (x, y) =√

x2 + y2 at (3, 4).

Solution. ∇f (x, y) = ( x√x2+y2

, y√x2+y2

), and ∇f (3, 4) = (3/5, 4/5),

so the linear approximation of f (x, y) at (3, 4) is given byL(x, y) = f (3, 4) +∇f (3, 4) · (x − 3, y − 4)= 5 + 3

5 (x − 3) + 45 (y − 4).

Then the number√(3.2)2 + (3.9)2 can be approximated by

L(3.2, 3.9) = 5 + 35 × 0.2 − 4

5 × 0.1 = 5 + 325 − 2

25 = 5.04.


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= f (a, b) +∇f (a, b) · (x − a, y − b)


Remark. For any fixed P(a, b), the function L(x, y) associated to afunction f at P(a, b) has a graph z = L(x, y) which is the tangent plane.

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x2 + y2 at (3, 4).


, y√x2+y2

), and ∇f (3, 4) = (3/5, 4/5),


5 (x − 3) + 45 (y − 4).


L(3.2, 3.9) = 5 + 35 × 0.2 − 4

5 × 0.1 = 5 + 325 − 2

25 = 5.04.


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= f (a, b) +∇f (a, b) · (x − a, y − b)



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x2 + y2 at (3, 4).


, y√x2+y2

), and ∇f (3, 4) = (3/5, 4/5),


5 (x − 3) + 45 (y − 4).


L(3.2, 3.9) = 5 + 35 × 0.2 − 4

5 × 0.1 = 5 + 325 − 2

25 = 5.04.


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= f (a, b) +∇f (a, b) · (x − a, y − b)



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x2 + y2 at (3, 4).


, y√x2+y2

), and ∇f (3, 4) = (3/5, 4/5),

so the linear approximation of f (x, y) at (3, 4) is given byL(x, y) = f (3, 4) +∇f (3, 4) · (x − 3, y − 4)

= 5 + 35 (x − 3) + 4

5 (y − 4).Then the number

√(3.2)2 + (3.9)2 can be approximated by

L(3.2, 3.9) = 5 + 35 × 0.2 − 4

5 × 0.1 = 5 + 325 − 2

25 = 5.04.


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= f (a, b) +∇f (a, b) · (x − a, y − b)



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x2 + y2 at (3, 4).


, y√x2+y2

), and ∇f (3, 4) = (3/5, 4/5),


5 (x − 3) + 45 (y − 4).


L(3.2, 3.9) = 5 + 35 × 0.2 − 4

5 × 0.1 = 5 + 325 − 2

25 = 5.04.


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Definition. In general, if f is a function defined in a domain D inRn (n = 2, 3), and P(x0) is a point in D, then the linear approximationof f at P is given by

L(x) = f (x0) +∇f (x0) · (x − x0),

= f (x0)

+ ∂f∂x1

(x0) · (x1 − a1) +∂f

∂x2(x0) · (x2 − a2) + · · ·+ ∂f

∂xn(x0) · (xn − an),

where ∇f (x0) = ( ∂f∂x1

(x0),∂f

∂x2(x0), · · · , ∂f

∂xn(x0) ), x0 = (a1, · · · , an),

and x = (x1, · · · , xn).

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Example. The function f (x, y) = 6x2 − 2x3 + y3 + 3y2 has n criticalpoints, then n = .

Solution. ∇f (x, y) = (12x − 6x2, 3y2 + 6y), ∇f (x, y) = (0, 0) if andonly 0 = 6x(x + 2) and 0 = 2y(y + 2), it follows that(x, y) = (0, 0), (−2, 0), (0,−2), (−2,−2) are the critical points of f .Then n = 4.


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The linear approximation of f (x, y) = x√

y at (1, 4) isA. 2 + 2x − y/4 B. 2 + 2x + y/4 C. 2 − 2x + y/4 D. 2x + y/4 − 1E. 2x − y/4 − 1

Solution. ∇f (x, y) = (√

y, x2√

y ), and ∇f (1, 4) = (2, 14 ),

hence the linear approximation of f at (1, 4) is given by

L(x, y) = f (1, 4) +∇f (1, 4) · (x − 1, y − 4)

= 2 + 2(x − 1) +14(y − 4)

= 2x +y4− 1.


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.Take-Home Quiz 3 (14 minutes)..

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...1 Explain the steps to determine the absolute extremum values ofa (differentiable) function z = f (x, y).

...2 State the second derivative test for critical points of z = f (x, y).

...3 Find the gradient ∇f of the function f (x, y, z) = 1√3−x2−y2−z2

for

at (−1, 1,−1).

Solution. ∇f = ∂f∂x i + ∂f

∂y j + ∂f∂x k = .

...4 Determine the minimum of the function f (x, y) = x2 + y2, whereR = {(x, y) | x2 + y2 + xy ≤ 1 }.Hint: 4x2 + 4y2 + 4xy = (x + 2y)2 + 3x2. For the boundary of theregion R, let x

√3 = cos t, and x + 2y = sin t for 0 ≤ t ≤ 2π,


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.Chain Rule I..

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If w = f (x, y, z) has continuous first-order partial derivatives and thatr(t) = (g(t), h(t), k(t)) is a differentiable curve in the domain of f , then

...1 the composite function F(t) = w ◦ r(t) = f (g(t), h(t), k(t)) is adifferentiable function of t, and

...2 its derivative is given bydFdt

=∂w∂x

· dxdt

+∂w∂y

· dydt

+∂w∂z

· dzdt

= ∇f (r(t)) · r′(t).

3

P(x¸, y¸, z¸)T

y

x

z

Remark. We omit the proof.


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.Chain Rule I..

......

If w = f (x, y, z) has continuous first-order partial derivatives and thatr(t) = (g(t), h(t), k(t)) is a differentiable curve in the domain of f , then

...1 the composite function F(t) = w ◦ r(t) = f (g(t), h(t), k(t)) is adifferentiable function of t, and

...2 its derivative is given bydFdt

=∂w∂x

· dxdt

+∂w∂y

· dydt

+∂w∂z

· dzdt

= ∇f (r(t)) · r′(t).

3

P(x¸, y¸, z¸)T

y

x

z

Remark. We omit the proof.Prepared by Dr. I.T. Leong for Tutorial 3-4 of Matb 210

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Suppose that x(t) = cos t, y(t) = sin t, z(t) = t andf (x, y, z) = exp(x2 + y2 + z2), find the derivative ofg(t) = f (x(t), y(t), z(t))with respect to t in terms of t only.A. 2et B. 2et2

C. 2e1+t2D. 2te1+t2

E. None of the above

Solution. Plugging the function into x, y, z components, we haveg(t) = f (x(t), y(t), z(t)) = f (sin t, cos t, t) = exp(cos2 t + sin2 t + t2) =

exp(1 + t2), so g′(t) = exp(1 + t2) ddt (1 + t2) = 2t exp(1 + t2).


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Theorem. Suppose that w = f (x, y, z) is a differentiable function,where x = x(u, v), y = y(u, v), z = z(u, v), where the coordinatefunctions are parameterized by differentiable functions. Then thecomposite function w(u, v) = f ( x(u, v), (u, v), (u, v) ) is adifferentiable function in u and v, such that the partial functions aregiven by

∂w∂u

=∂w∂x

· ∂x∂u

+∂w∂y

· ∂y∂u

+∂w∂z

· ∂z∂u

;

∂w∂v

=∂w∂x

· ∂x∂v

+∂w∂y

· ∂y∂v

+∂w∂z

· ∂z∂v

.

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Remark. The formula stated above is very important in the theory ofsurface integral.


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.Chain Rules of Composite Functions..

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Theorem. Suppose that s = f (x, y, z) is a differentiable function,where x = x(u, v, w), y = y(u, v, w), z = z(u, v, w) are thecoordinate functions parameterized by differentiable functions invariables u, v and w. Then the composite functionS(u, v, w) = f ( x(u, v, w), (u, v, w), (u, v, w) ) is differentiable in u, vand w, such that the partial functions are given by

∂S∂u

=∂S∂x

· ∂x∂u

+∂S∂y

· ∂y∂u

+∂S∂z

· ∂z∂u

;

∂S∂v

=∂S∂x

· ∂x∂v

+∂S∂y

· ∂y∂v

+∂S∂z

· ∂z∂v

;

∂S∂w

=∂S∂x

· ∂x∂w

+∂S∂y

· ∂y∂w

+∂S∂z

· ∂z∂w

.

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Remark. The formula stated above is very important in the theory ofinverse function theory and integration theory.


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Example. (a) Let f (x, y) = (ex+y, ex−y). Let r(t) be a path withr(0) = (0, 0) and r′(0) = (1, 1). What is the tangent vector to theimage of r(t) under f at t = 0?

Solution. Let r(t) = (x(t), y(t)). It follows from the given condition thatx(0) = y(0) = 0, and x′(0) = y′(0) = 1. By composition, we haves(t) = (u(t), v(t)) = f ◦ r(t) = (ex(t)+y(t), ex(t)−y(t)). Then it follows

from the chain rule thatddt

u(t) =ddt(ex(t)+y(t))

= ex(t)+y(t) ddt(x(t) + y(t)) = ex(t)+y(t)(x′(t) + y′(t)).

In particular,dudt

|t=0 = e0+0 · (1 + 1) = 2, similarly,dvdt

|t=0 = e0−0 · (1 − 1) = 0. Hence the curve s(t) = f ◦ r(t) has a

tangent vector s′(0) = (2, 0) at t = 0.


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Example. Let w = f (x, y), where x = x(r, θ) = r cos θ andy = y(r, θ) = r sin θ. Consider the composite functionw = f (r cos θ, r sin θ) as a function of variables r and θ.Express ∂w

∂x , ∂w∂y and ( ∂w

∂x )2 + ( ∂w

∂y )2 in terms of ∂w

∂r and ∂w∂θ .

Solution. It follows from previous problem that1r ·

∂w∂θ = sin θ fx(r cos θ, r sin θ)− cos θ fy(r cos θ, r sin θ), and∂w∂r = cos θ fx(r cos θ, r sin θ) + sin θ fy(r cos θ, r sin θ).

Now think of a system of linear equations in unknown fx and fy:{wθr = sin θ · fx − cos θ · fy, (1)

wr = cos θ · fx + sin θ · fy. (2)Eliminate fy from (1) and (2) by considering sin θ × (1) + cos θ × (2),so

sin θ · wθr + cos θ · wr = sin2 θ fx + cos2 θ fx = fx.

Eliminate fx from (1) and (2) by considering− cos θ × (1) + sin θ × (2), so

− cos θ · wθr + sin θ · wr = cos2 θ fy + sin2 θ fy = fy, and

f 2x + f 2

y =(sin θ · wθ

r + cos θ · wr)2

+(− cos θ · wθ

r + sin θ · wr)2

= (sin2 θ + cos2 θ)

(w2

θ

r2 + w2r

)=

w2θ

r2 + w2r .Prepared by Dr. I.T. Leong for Tutorial 3-4 of Matb 210

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Suppose that z = f (x, y) is a function defined in a domain D, andP(a, b) is a point in D. Recall that the partial derivatives

fx(a, b) = limh→0

f (a + h, b)− f (a, b)h

, and

fy(a, b) = limk→0

f (a, b + k)− f (a, b)k

.

The limits are taken along the coordinate axes.


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Definition (Directional derivative) The resulting derivative g′(0) iscalled the directional derivative Duf of f along the direction u, andhence we write Duf = ∇f · u to represent the rate of the change of fin the unit direction u.

Remark. In general, if f = f (x1, · · · , xn) is a function of n variables,one define(i) the gradient of f to be ∇f = ( ∂f

∂x1, · · · , ∂f

∂xn), and

(ii) the directional derivative Duf by Duf = ∇f · u.


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Example. Find the directional derivative of f (x, y) = tan(x + 2y) at thepoint (0, π/6) in the direction of (−3, 4).

Solution. The unit direction vector is u =(−3, 4)√33 + 42

= (−35

,45).

Moreover, ∇f (x, y) =∇(x + 2y)

cos2(x + 2y)=

i + 2jcos2(x + 2y)

, so

∇f (0, π/6) = 4(i + 2j). Hence

Duf (0, π/6) = ∇f (0, π/6) · u = (−35)× 4+ (

85)× 4 =

−12 + 325

= 4.

Remark. The direction vector u should be of unit length, i.e. ∥u∥ = 1.


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Proposition. The greatest rate of change of a scalar function f , i.e.,the maximum directional derivative, takes place in the direction of,and has the magnitude of, the vector ∇f .

Proof. For any direction v, the directional derivative of f along thedirection v at a point P in the domain of f , is given byDv(P) := ⟨∇f (P), v

∥v∥ ⟩ = ∥∇f∥ cos θ, where θ is the angle betweenthe vectors ∇f (P) and v. Hence Dv(P) attains maximum (minimum)value if and only if cos θ = 1 (−1), if and only if ∇f (P) ( −∇f (P) ) isparallel to v. In this case, we have Dv(P) = ∥∇f∥ ( −∥∇f∥ ).


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Example. (a) Find the directional derivative of f (x, y, z) = 2x3y − 3y2zat P(1, 2,−1) in a direction v = 2i − 3j + 6k.(b) In what direction from P is the directional derivative a maximum?(c) What is the magnitude of the maximum directional derivative?

Solution.(a) ∇f (P) = 6x2yi + (2x3 − 6yz)j − 3y2k|(1,2,−1) = 12i + 14j − 12k at

P. Then the directional derivative of f along the direction v isgiven by Dvf = ∇f · v

∥v∥ = ⟨12i + 14j − 12k, 2i−3j+6k√22+32+62 ⟩ = − 90

7 .(b) Dvf (P) is maximum(minimum) ⇐⇒ v (−v) is parallel to

∇f (P) = 12i + 14j − 12k.(c) The maximum magnitude of Dvf (P) is given by

∇f · ∇f (P)∥∇f (P)∥ = ∥∇f (P)∥2

∥∇f (P)∥ = ∥∇f (P)∥ = ∥12i + 14j − 12k∥ =√

144 + 196 + 144 = 22.


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Example. The temperature in the vicinity of a polar bear is given byT(x, y, z) = e−x + e−2y + e3z. If it is at (1, 1, 1), in what direction shouldthe bear proceed in order to cool fastest?

Solution.

In order to cool the fastest, the bear should proceed in thedirection in which T is decreasing the fastest; that is, in the direction−∇T(1, 1, 1). As∇T(x, y, z) = ∂T

∂x i + ∂T∂y j + ∂T

∂z k = (−e−x,−2e−2y, 3e3x). Thus the

required direction is −∇T(1, 1, 1) = (1/e, 2/e2,−3e3)..

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Example. Let u = f (x, y, z) = e−x2sin(xy). In what direction from

(1, π, 0) should one proceed to increase f most rapidly?

Solution. ∇f (x, y, z) = ∂u∂x i + ∂u

∂y j + ∂u∂x k =

y cos(xy)e−z2i + x cos(xy)e−z2

j − 2z sin(xy)e−z2k, and hence

∇f (1, π, 0) = (−π,−1, 0). So the required direction is the one givenby (−π,−1, 0).


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Solution. In order to cool the fastest, the bear should proceed in thedirection in which T is decreasing the fastest; that is, in the direction−∇T(1, 1, 1). As∇T(x, y, z) = ∂T

∂x i + ∂T∂y j + ∂T


required direction is −∇T(1, 1, 1) = (1/e, 2/e2,−3e3).

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∂y j + ∂u∂x k =





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∂x i + ∂T∂y j + ∂T



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Solution.

∇f (x, y, z) = ∂u∂x i + ∂u

∂y j + ∂u∂x k =





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∂x i + ∂T∂y j + ∂T



......




∂y j + ∂u∂x k =





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Example. Find the rate of change of f (x, y) = 2x3 − xy + xy2 at (2, 3)in the direction of 3i − 4j.A. 5 B. 10 C. 50 D. None of the above.

Solution.

Take u = 3i−4j∥3i−4j∥ = 3i/5 − 4j/5 = (3/5,−4/5). And

∇f (x, y) = (6x2 − y + y2,−x + 2xy). Then the directional derivativeDuf (2, 3) = ∇f (2, 3) · u = 10. Hence the correct answer is B.


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Example. Find the rate of change of f (x, y) = 2x3 − xy + xy2 at (2, 3)in the direction of 3i − 4j.A. 5 B. 10 C. 50 D. None of the above.

Solution. Take u = 3i−4j∥3i−4j∥ = 3i/5 − 4j/5 = (3/5,−4/5). And

∇f (x, y) = (6x2 − y + y2,−x + 2xy). Then the directional derivativeDuf (2, 3) = ∇f (2, 3) · u = 10. Hence the correct answer is B.


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Proposition. Let f (x, y, z) be a differentiable function defined in R3,and S : f (x, y, z) = c be a level surface for some constant c, i.e.S = { (x, y, z) | f (x, y, z) = c }. Suppose that P(a, b, c) be a point on Ssuch that the gradient vector ∇f (a, b, c) of f at point P(a, b, c) is notzero, then the equation of the tangent plane of S at P is given by< ∇f (a, b, c), (x − a, y − b, z − c) >= 0, i.e.

fx(a, b, c)(x − a) + fy(a, b, c)(y − b) + fz(a, b, c)(z − c) = 0.

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Remark. For any given level surface S defined by a scalar function f ,the tangent plane of S at any P of S is spanned by the tangent vectorof the curve contained in S. The result above tells us that the normaldirection to the tangent plane of S at any point P of S is parallel to∇f (P).


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Example. Let n be the number points on the level surface (ellipsoid)f (x, y, z) = x2 + 2y2 + 4z2 + xy + 3yz = 1 at which the tangent plane isparallel to the xz-plane. Then n =

A. 0 B. 1 C. 2 D. None of the above.

Solution.

We first evaluate the gradient of f , then∇f (x, y, z) = ( 2x + y, 4y + x + 3z, 8z + 3y ).If the tangent plane of level surface given by f (x, y, z) = 1 is parallel tothe xz-plane, then ∇f is parallel to (0, 1, 0). In particular, we have(x, y, z) must satisfies y = 2x, 3y = 8z and the defining equationx2 + 2y2 + 4z2 + xy + 3yz = 1. After multiplying the last equation by16, we have 32 = 4(2x)2 + 32y2 + (8z)2 + 8(2x)y + 24(y)(8z)= 4y2 + 32y2 + (3y)2 + 8y2 + 72y2 = 125y2. So y = ±

√32/125, from

these we know the corresponding x and y coordinates.In particular, there are only two points on the level surface satisfyingthe required condition.


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Solution. We first evaluate the gradient of f , then∇f (x, y, z) = ( 2x + y, 4y + x + 3z, 8z + 3y ).

If the tangent plane of level surface given by f (x, y, z) = 1 is parallel tothe xz-plane, then ∇f is parallel to (0, 1, 0). In particular, we have(x, y, z) must satisfies y = 2x, 3y = 8z and the defining equationx2 + 2y2 + 4z2 + xy + 3yz = 1. After multiplying the last equation by16, we have 32 = 4(2x)2 + 32y2 + (8z)2 + 8(2x)y + 24(y)(8z)= 4y2 + 32y2 + (3y)2 + 8y2 + 72y2 = 125y2. So y = ±

√32/125, from



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Solution. We first evaluate the gradient of f , then∇f (x, y, z) = ( 2x + y, 4y + x + 3z, 8z + 3y ).If the tangent plane of level surface given by f (x, y, z) = 1 is parallel tothe xz-plane, then ∇f is parallel to (0, 1, 0).

In particular, we have(x, y, z) must satisfies y = 2x, 3y = 8z and the defining equationx2 + 2y2 + 4z2 + xy + 3yz = 1. After multiplying the last equation by16, we have 32 = 4(2x)2 + 32y2 + (8z)2 + 8(2x)y + 24(y)(8z)= 4y2 + 32y2 + (3y)2 + 8y2 + 72y2 = 125y2. So y = ±

√32/125, from



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Solution. We first evaluate the gradient of f , then∇f (x, y, z) = ( 2x + y, 4y + x + 3z, 8z + 3y ).If the tangent plane of level surface given by f (x, y, z) = 1 is parallel tothe xz-plane, then ∇f is parallel to (0, 1, 0). In particular, we have(x, y, z) must satisfies y = 2x, 3y = 8z and the defining equationx2 + 2y2 + 4z2 + xy + 3yz = 1.

After multiplying the last equation by16, we have 32 = 4(2x)2 + 32y2 + (8z)2 + 8(2x)y + 24(y)(8z)= 4y2 + 32y2 + (3y)2 + 8y2 + 72y2 = 125y2. So y = ±

√32/125, from



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Solution. We first evaluate the gradient of f , then∇f (x, y, z) = ( 2x + y, 4y + x + 3z, 8z + 3y ).If the tangent plane of level surface given by f (x, y, z) = 1 is parallel tothe xz-plane, then ∇f is parallel to (0, 1, 0). In particular, we have(x, y, z) must satisfies y = 2x, 3y = 8z and the defining equationx2 + 2y2 + 4z2 + xy + 3yz = 1. After multiplying the last equation by16, we have 32 = 4(2x)2 + 32y2 + (8z)2 + 8(2x)y + 24(y)(8z)= 4y2 + 32y2 + (3y)2 + 8y2 + 72y2 = 125y2. So y = ±

√32/125, from

these we know the corresponding x and y coordinates.

In particular, there are only two points on the level surface satisfyingthe required condition.


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Solution. We first evaluate the gradient of f , then∇f (x, y, z) = ( 2x + y, 4y + x + 3z, 8z + 3y ).If the tangent plane of level surface given by f (x, y, z) = 1 is parallel tothe xz-plane, then ∇f is parallel to (0, 1, 0). In particular, we have(x, y, z) must satisfies y = 2x, 3y = 8z and the defining equationx2 + 2y2 + 4z2 + xy + 3yz = 1. After multiplying the last equation by16, we have 32 = 4(2x)2 + 32y2 + (8z)2 + 8(2x)y + 24(y)(8z)= 4y2 + 32y2 + (3y)2 + 8y2 + 72y2 = 125y2. So y = ±

√32/125, from



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Example. Show that the surface S : x2 − 2yz + y3 = 4 is perpendicularto any member of the family of surfaces Sa : x2 + 1 = (2 − 4a)y2 + az2

at the point of intersection P(1,−1, 2).

Solution.

Let the defining equations of level surfaces S, Sa beF(x, y, z) = x2 − 2yz + y3 − 4 = 0 andG(x, y, z) = x2 + 1 − (2 − 4a)y2 − az2 = 0. Then∇F(x, y, z) = 2xi + (3y2 − 2z)j − 2yk, and∇G(x, y, z) = 2xi − 2(2 − 4a)yj − 2azk.

Thus, the normals to the two surfaces at P(1,−1, 2) are given byN1 = 2i − j + 2k, N2 = 2i + 2(2 − 4a)j − 4ak.Since N1 · N2 = (2)(2)− 2(2 − 4a)− (2)(4a) = 0, it follows that N1and N2 are perpendicular for all a, and so the required result follows.


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Solution. Let the defining equations of level surfaces S, Sa beF(x, y, z) = x2 − 2yz + y3 − 4 = 0 andG(x, y, z) = x2 + 1 − (2 − 4a)y2 − az2 = 0. Then∇F(x, y, z) = 2xi + (3y2 − 2z)j − 2yk, and∇G(x, y, z) = 2xi − 2(2 − 4a)yj − 2azk.



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Thus, the normals to the two surfaces at P(1,−1, 2) are given byN1 = 2i − j + 2k, N2 = 2i + 2(2 − 4a)j − 4ak.

Since N1 · N2 = (2)(2)− 2(2 − 4a)− (2)(4a) = 0, it follows that N1and N2 are perpendicular for all a, and so the required result follows.


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Let f (x, y) be a differentiable function defined on xy-plane. For anyreal number k, recall the level level of f for k is given by the set{ (x, y) | f (x, y) = k }. When the value k changes, the level curvechanges gradually.

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Let f (x, y) = x2 − 7xy + 2y2 defined onxy-plane. The blue curves represent the levelcurves Ck : f (x, y) = k of various values c. Andthe red arrows represent the gradient vectorfield ∇f (a, b) = ( fx(a, b), fy(a, b) ) which isnormal to the tangent vector to level curve Caat P(a, b) of various values k.


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Proposition. Let Ck : f (x, y) = k be a fixed level curve with a pointP(a, b) in C − k. If ∇f (a, b) ̸= (0, 0), then the equation of the tangentline of Ck at P is given by ∇f (a, b) · (x − a, y − b) = 0, i.e.fx(a, b)(x − a) + fy(a, b)(y − b) = 0.

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Example. Let n be the number of points in the level curveC : 3x2 − 4y2 = −1 at which the the normal vector of C isperpendicular to the vector 4i + 3j. Then n =


Solution. ∇(3x2 − 4y2) = (6x,−8y). Then ∇f ⊥ 4i + 3j if and only if24x − 24y = 0 i.e. x = y. Then −1 = 3x2 − 4x2 = −x2. Hence thereare only two points, namely (1, 1) and (−1,−1), lying on the givencurve satisfying the condition.


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Solution.

∇(3x2 − 4y2) = (6x,−8y). Then ∇f ⊥ 4i + 3j if and only if24x − 24y = 0 i.e. x = y. Then −1 = 3x2 − 4x2 = −x2. Hence thereare only two points, namely (1, 1) and (−1,−1), lying on the givencurve satisfying the condition.


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Solution. ∇(3x2 − 4y2) = (6x,−8y). Then ∇f ⊥ 4i + 3j if and only if24x − 24y = 0 i.e. x = y. Then −1 = 3x2 − 4x2 = −x2. Hence thereare only two points, namely (1, 1) and (−1,−1), lying on the givencurve satisfying the condition.


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Implicit Function Theorem II.If the level surface (or implicit function) S : F(x, y, z) = c defines anexplicit function z = z(x, y), and P ∈ S.

If Fz(P) ̸= 0, then∂z∂x

= −Fx

Fzand

∂z∂y

= −Fy

Fz.

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Remark. In fact, Implicit Function Theorem tells us more. Iffz(a, b, c) ̸= 0, then there exists a function z = g(x, y) defined on asmall ball B(Q, δ) of Q(a, b) in xy-plane such that all the points on thelevel surface S : f (x, y, z) = f (a, b, c) is given by the graph of f , i.e.f (x, g(x, y)) = f (a, b, c). In other words, the level surface locally lookslike a deformed sheet of xy-plane. More generally, if ∇f (a, b, c) is nota zero vector, then one can choose a right projection, so that locallythe level surface S is a graph given by a differentiable function of 2variables. In particular, it is locally a smooth surface with a tangentplane. This is why level surface can be thought as an implicit function,in which one can use it to define an implicit function in which one ofthe variable is expressed in terms of the other 2 variables.


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= −Fx

Fzand

∂z∂y

= −Fy

Fz.

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Remark. In fact, Implicit Function Theorem tells us more. Iffz(a, b, c) ̸= 0, then there exists a function z = g(x, y) defined on asmall ball B(Q, δ) of Q(a, b) in xy-plane such that all the points on thelevel surface S : f (x, y, z) = f (a, b, c) is given by the graph of f , i.e.f (x, g(x, y)) = f (a, b, c). In other words, the level surface locally lookslike a deformed sheet of xy-plane.

More generally, if ∇f (a, b, c) is nota zero vector, then one can choose a right projection, so that locallythe level surface S is a graph given by a differentiable function of 2variables. In particular, it is locally a smooth surface with a tangentplane. This is why level surface can be thought as an implicit function,in which one can use it to define an implicit function in which one ofthe variable is expressed in terms of the other 2 variables.


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= −Fx

Fzand

∂z∂y

= −Fy

Fz.

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Remark. In fact, Implicit Function Theorem tells us more. Iffz(a, b, c) ̸= 0, then there exists a function z = g(x, y) defined on asmall ball B(Q, δ) of Q(a, b) in xy-plane such that all the points on thelevel surface S : f (x, y, z) = f (a, b, c) is given by the graph of f , i.e.f (x, g(x, y)) = f (a, b, c). In other words, the level surface locally lookslike a deformed sheet of xy-plane. More generally, if ∇f (a, b, c) is nota zero vector, then one can choose a right projection, so that locallythe level surface S is a graph given by a differentiable function of 2variables. In particular, it is locally a smooth surface with a tangentplane. This is why level surface can be thought as an implicit function,in which one can use it to define an implicit function in which one ofthe variable is expressed in terms of the other 2 variables.


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Example. Let z(x, y) be defined by means of an implicit functionx2 + y2 + z2 = 3xyz. Prove that ∂z

∂x = 3yz−2x2z−3xy , and ∂z

∂y = 3xz−2y2z−3xy .

Proof. Replace the variable z in the implicit function by an explicitfunction z = z(x, y), so we have x2 + y2 + (z(x, y))2 = 3xy · z(x, y) ♡in a neighborhood of the point (x, y) = (a, b) of the domain z = z(x, y).

Differentiate the identity ♡ with respect to the independent variables xand y respectively, so that we have2x + 2z(x, y) · ∂z

∂x = 3y · z(x, y) + 3xy · ∂z∂x ,

then we can collect the term ∂z∂x , so

∂z∂x (2z(x, y)− 3xy) = 3yz(x, y)− 2x, i.e. ∂z

∂x = 3yz(x,y)−2x2z(x,y)−3xy .

Remark. We usually ignore the dependent of z with respect to x andy, because it is clear that ∂z

∂x means that z is a function of on x and theremaining independent variables.We skip the details for ∂z

∂y .


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∂y = 3xz−2y2z−3xy .

Proof. Replace the variable z in the implicit function by an explicitfunction z = z(x, y), so we have x2 + y2 + (z(x, y))2 = 3xy · z(x, y) ♡in a neighborhood of the point (x, y) = (a, b) of the domain z = z(x, y).Differentiate the identity ♡ with respect to the independent variables xand y respectively, so that we have2x + 2z(x, y) · ∂z

∂x = 3y · z(x, y) + 3xy · ∂z∂x ,



∂x = 3yz(x,y)−2x2z(x,y)−3xy .



∂y .


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∂y = 3xz−2y2z−3xy .


∂x = 3y · z(x, y) + 3xy · ∂z∂x ,



∂x = 3yz(x,y)−2x2z(x,y)−3xy .


∂x means that z is a function of on x and theremaining independent variables.

We skip the details for ∂z∂y .


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∂y = 3xz−2y2z−3xy .


∂x = 3y · z(x, y) + 3xy · ∂z∂x ,



∂x = 3yz(x,y)−2x2z(x,y)−3xy .



∂y .


. . . . . .

.

......

Example. Determine the extremum of the function z = z(x, y) definedimplicitly by the equation 3x2 + 2y2 + z2 + 8yz − z + 8 = 0.

Solution. Define F(x, y, z) = 3x2 + 2y2 + z2 + 8yz − z + 8, so thefunction z = z(x, y) is in fact the graph of the level surface Sassociated to the equation F(x, y, z) = 0, or sometimes we just denoteit by S : F(x, y, z) = 0.

It follows that F(x, y, z(x, y)) = 0, for all (x, y) inthe (unspecified) domain of z(x, y), in fact we just think of the equalityas an identity in x and y. So differentiate with respect to x and yrespectively by means of chain rule, we have

0 =∂

∂x(0) =

∂

∂x( F(x, y, z(x, y)) ) =

∂F∂x

∂x∂x

+∂F∂z

∂z∂x

= Fx + Fz∂z∂x

, so

one has∂z∂x

(x, y) = −Fx(x, y, z(x, y))Fz(x, y, z(x, y))

= −Fx

Fzand

∂z∂y

(x, y) = −Fy(x, y, z(x, y))Fz(x, y, z(x, y))

= −Fy

Fz. One should notice that the

assumption Fz ̸= 0 for all (x, y) in the domain of z = z(x, y) isnecessary, which one can obtain explicitly if Fz is known.


. . . . . .

.

......


Solution. Define F(x, y, z) = 3x2 + 2y2 + z2 + 8yz − z + 8, so thefunction z = z(x, y) is in fact the graph of the level surface Sassociated to the equation F(x, y, z) = 0, or sometimes we just denoteit by S : F(x, y, z) = 0. It follows that F(x, y, z(x, y)) = 0, for all (x, y) inthe (unspecified) domain of z(x, y), in fact we just think of the equalityas an identity in x and y.

So differentiate with respect to x and yrespectively by means of chain rule, we have

0 =∂

∂x(0) =

∂

∂x( F(x, y, z(x, y)) ) =

∂F∂x

∂x∂x

+∂F∂z

∂z∂x

= Fx + Fz∂z∂x

, so

one has∂z∂x


= −Fx

Fzand

∂z∂y


= −Fy




. . . . . .

.

......


Solution. Define F(x, y, z) = 3x2 + 2y2 + z2 + 8yz − z + 8, so thefunction z = z(x, y) is in fact the graph of the level surface Sassociated to the equation F(x, y, z) = 0, or sometimes we just denoteit by S : F(x, y, z) = 0. It follows that F(x, y, z(x, y)) = 0, for all (x, y) inthe (unspecified) domain of z(x, y), in fact we just think of the equalityas an identity in x and y. So differentiate with respect to x and yrespectively by means of chain rule, we have

0 =∂

∂x(0) =

∂

∂x( F(x, y, z(x, y)) ) =

∂F∂x

∂x∂x

+∂F∂z

∂z∂x

= Fx + Fz∂z∂x

, so

one has∂z∂x


= −Fx

Fzand

∂z∂y


= −Fy




. . . . . .

.

......


Solution. Let F(x, y, z) = 3x2 + 2y2 + z2 + 8yz − z + 8.

so∂z∂x

= −Fx

Fz= − 6x

2z + 8y − 1, and

∂z∂y

= −Fy

Fz= − 4y + 8z

2z + 8y − 1. To

locate the extremum value of z = z(x, y), one need its two partialderivatives zx and zy vanish, i.e. (6x, 4y + 8z) = (0, 0) where (x, y, z)is a point of the level surface S : F(x, y, z) = 0. Hence, x = 0, andy = −2z. Then 0 = F(0,−2z, z) = 2(−2z)2 + z2 + 8(−2z)z − z + 8 =

−7z2 − z + 8 = −(7z + 8)(z − 1) so z = 1 or − 87 . Hence P(0,−2, 1) or

Q(0, 167 ,− 8

7 ) are the only critical point of the function z = z(x, y),however, z = z(x, y) is not explicitly determined yet. One candetermine use the quadratic formula to express z in terms of x and y,and then one can see that zmax = 1 and zmin = − 8

7 .Remark. In the last part, we skip some details, but the gap can befilled in after we learn the second derivative test.


. . . . . .

.

......


Solution. Let F(x, y, z) = 3x2 + 2y2 + z2 + 8yz − z + 8. so∂z∂x

= −Fx

Fz= − 6x

2z + 8y − 1, and

∂z∂y

= −Fy

Fz= − 4y + 8z

2z + 8y − 1.

To



Q(0, 167 ,− 8




. . . . . .

.

......



= −Fx

Fz= − 6x

2z + 8y − 1, and

∂z∂y

= −Fy

Fz= − 4y + 8z

2z + 8y − 1. To

locate the extremum value of z = z(x, y), one need its two partialderivatives zx and zy vanish, i.e.

(6x, 4y + 8z) = (0, 0) where (x, y, z)is a point of the level surface S : F(x, y, z) = 0. Hence, x = 0, andy = −2z. Then 0 = F(0,−2z, z) = 2(−2z)2 + z2 + 8(−2z)z − z + 8 =


Q(0, 167 ,− 8




. . . . . .

.

......



= −Fx

Fz= − 6x

2z + 8y − 1, and

∂z∂y

= −Fy

Fz= − 4y + 8z

2z + 8y − 1. To

locate the extremum value of z = z(x, y), one need its two partialderivatives zx and zy vanish, i.e. (6x, 4y + 8z) = (0, 0) where (x, y, z)is a point of the level surface S : F(x, y, z) = 0. Hence, x = 0, andy = −2z.

Then 0 = F(0,−2z, z) = 2(−2z)2 + z2 + 8(−2z)z − z + 8 =


Q(0, 167 ,− 8




. . . . . .

.

......



= −Fx

Fz= − 6x

2z + 8y − 1, and

∂z∂y

= −Fy

Fz= − 4y + 8z

2z + 8y − 1. To


−7z2 − z + 8 = −(7z + 8)(z − 1)

so z = 1 or − 87 . Hence P(0,−2, 1) or

Q(0, 167 ,− 8




. . . . . .

.

......



= −Fx

Fz= − 6x

2z + 8y − 1, and

∂z∂y

= −Fy

Fz= − 4y + 8z

2z + 8y − 1. To



Q(0, 167 ,− 8




. . . . . .

.

......

Example. Find second order partial derivative ∂2

∂y∂x z(x, y) of thefunction z = z(x, y) defined implicitly by the equation3x2 + 2y2 + z2 + 8yz − z + 8 = 0.

Solution. We follow the notation of the previous problem. In the last

problem, we have∂z∂x

= − 6x2z + 8y − 1

, and∂z∂y

= − 4y + 8z2z + 8y − 1

. So

we have zxy =∂zx

∂y=

∂

∂y

(− 6x(2z + 8y − 1)2

)=

− 6x(2z + 8y − 1)2 · ∂

∂y(2z + 8y − 1) = − 6x

(2z + 8y − 1)2 · (2zx + 8) =

− 6x(2z + 8y − 1)2 · (8 − 12x

2z + 8y − 1).


. . . . . .

.

......

If the plane π : lx + my + nz = p and the quadric surfaceS : Ax2 + By2 + Cz2 = 1 are tangent to each other, determine therelationship among the coefficients.

Solution. The necessary and sufficient condition is l2A + m2

B + n2

C = p2.Note that both surfaces π and S are level surfaces defined byfunctions f (x, y, z) = lx + my + nz and g(x, y, z) = Ax2 + By2 + Cz2 Atthe common point P(x, y, z) (if exists) of two surfaces, both surfaceswill be tangent to each other, i.e. they have the same normal vectors,hence ∇f (x, y, z) = (l, m, n) and ∇g(x, y, z) = (2xA, 2yB, 2zC) areparallel, so (xA, yB, zC) = λ(l, m, n) for some non-zero λ, i.e.(x, y, z) = λ( l

A , mB , n

C ). As the coordinates of the common point

P(x, y, z) satisfy both defining equations, so p = λ(

l2A + m2

B + n2

C

)and

1 = λ2(

A l2A2 + B m2

B2 + C n2

C2

)= λ2

(l2A + m2

B + n2

C

)= λp, i.e.

p2 = p · p = pλ

(l2

A+

m2

B+

n2

C

)=

l2

A+

m2

B+

n2

C.


. . . . . .

.

......

If the plane π : lx + my + nz = p and the quadric surfaceS : Ax2 + By2 + Cz2 = 1 are tangent to each other, determine therelationship among the coefficients.

Solution. The necessary and sufficient condition is l2A + m2

B + n2

C = p2.On the other hand, suppose that the coefficients satisfy the equation,let P(x0, y0, z0) = P( l

Ap , mpB , n

Cp ), then

lx0 + my0 + nz0 =l2

Ap+

m2

pB+

n2

Cp=

1p

(l2

A+

m2

B+

n2

C

)=

p2

p= p,

i.e. P lies on π. And Ax20 + By2

0 + Cz20 = A l2

A2p2 + B m2

p2B2 + C n2

C2p2 =

1p2

(l2A + m2

B + n2

C

)= p2

p2 = 1, i.e. P lies on S too. So P lies on the

intersection of surfaces S and T. One then check ∇g = 2p∇f , and so

these are tangent to each at P.


. . . . . .

.

......

Example. Suppose that the variables P, V and T satisfy the equationof famous ideal gas law: PV = nRT, where n and R are constantsindependent of P, V and T. If P = P(V, T), V = V(P, T), andT = T(P, V) are implicit functions defined by means of the levelsurface S : PV = nRT.

(i) Evaluate the partial derivatives:∂P∂T

(T, V),∂T∂V

(P, V) and∂V∂P

(P, V).

(ii) Simplify∂P∂T

· ∂T∂V

· ∂V∂P

.

Solution. (i) As P = nRTV , so

∂P∂T

(T, V) =nRV

. Similarly T = PVnR , and

hence∂T∂V

(P, V) =P

nR. And V = nRT

P , so∂V∂P

(P, V) = −nRTP2 .

(ii)∂P∂T

· ∂T∂V

· ∂V∂P

= −nRV

· PnR

· .nRTP2 = −nRT

PV= −1.


. . . . . .

.Quiz 4 (14 minutes)..

......

...1 State the chain rule for the composite function of continuouslydifferentiable function f (x, y, z) and a differentiable curver(t) = (x(t), y(t), z(t)) in space.

...2 Suppose that x = r cos θ, y = r sin θ, and let z = f (x, y) be a twicecontinuously differentiable function. Defineg(r, θ) = f (r cos θ, , r sin θ). Express the partial derivatives:

(i) gθ(r, θ) =∂g∂θ

= , and

(ii)∂2g∂θ2 = ,

i.e. gθθ in terms of the partial derivatives of f ....3 Find the directional derivative Duf = of

f (x, y, z) =√

x2 + y2 − z2 at the point P(1, 1, 1) along the unitdirection u parallel to (1, 1,−1).


tutorial class iii and iv 17-29/09/2012 1. first order...

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