chainrule_implicit differentiation

23
Chapter 7 The Chain Rule, Related Rates, and Implicit Differentiation 7.1 For each of the following, find the derivative of y with respect to x. (a) y 6 +3y - 2x - 7x 3 =0 (b) e y +2xy = 3 (c) y = x cos x Detailed Solution: (a) d dx (y 6 +3y - 2x - 7x 3 ) = d dx (0) 6y 5 dy dx +3 dy dx - 2 - 21x 2 = 0 (6y 5 + 3) dy dx = 21x 2 +2 dy dx = 21x 2 +2 6y 5 +3 (b) d dx (e y +2xy ) = d dx ( 3) e y dy dx + (2y +2x dy dx ) = 0 (e y +2x) dy dx = -2y dy dx = - 2y e y +2x v.2005.1 - September 4, 2009 1

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Page 1: chainRule_implicit differentiation

Chapter 7

The Chain Rule, Related Rates, and

Implicit Differentiation

7.1

For each of the following, find the derivative of y with respect to x.

(a) y6 + 3y − 2x − 7x3 = 0

(b) ey + 2xy =√

3

(c) y = xcos x

Detailed Solution:

(a)

d

dx(y6 + 3y − 2x − 7x3) =

d

dx(0)

6y5dy

dx+ 3

dy

dx− 2 − 21x2 = 0

(6y5 + 3)dy

dx= 21x2 + 2

dy

dx=

21x2 + 2

6y5 + 3

(b)

d

dx(ey + 2xy) =

d

dx(√

3)

ey dy

dx+ (2y + 2x

dy

dx) = 0

(ey + 2x)dy

dx= −2y

dy

dx= −

2y

ey + 2x

v.2005.1 - September 4, 2009 1

Page 2: chainRule_implicit differentiation

Math 102 Problems Chapter 7

(c) First take the logarithm of both sides: ln y = ln(xcos x) = (cos x)(ln x), then differentiate bothsides with respect to x.

d

dx(ln y) =

d

dx(cos x · ln x)

1

y

dy

dx= − sin x · ln x + cos x ·

1

xdy

dx= y

(

− sin x · ln x +cos x

x

)

= xcos x(

− sin x · ln x +cos x

x

)

7.2

Consider the growth of a cell, assumed spherical in shape. Suppose that the radius of the cellincreases at a constant rate per unit time. (Call the constant k, and assume that k > 0.)

(a) At what rate would the volume, V , increase ?

(b) At what rate would the surface area, S, increase ?

(c) At what rate would the ratio of surface area to volume S/V change? Would this ratio increaseor decrease as the cell grows? [Remark: note that the answers you give will be expressed interms of the radius of the cell.]

Detailed Solution:

We are given the information thatdr

dt= k.

(a) For the volume:

V =4

3πr3

dV

dt=

3

d(r3)

dt=

33r2

dr

dt= 4πr2k.

(b) For the surface area:S = 4πr2

dS

dt=

d

dt(4πr2) = 4π(2r)

dr

dt= 8πrk.

(c)S

V=

4πr2

(4/3)πr3=

3

r

Thusd

dt

(

S

V

)

=d

dt

(

3

r

)

= −3

r2

dr

dt= −

3k

r2.

The derivative is negative so that the ratio of surface area to volume is decreasing as the cellgrows.

v.2005.1 - September 4, 2009 2

Page 3: chainRule_implicit differentiation

Math 102 Problems Chapter 7

7.3 Growth of a circular fungal colony

A fungal colony grows on a flat surface starting with a single spore. The shape of the colony edgeis circular (with the initial site of the spore at the center of the circle.) Suppose the radius of thecolony increases at a constant rate per unit time. (Call this constant C.)

(a) At what rate does the area covered by the colony change ?

(b) The biomass of the colony is proportional to the area it occupies (factor of proportionalityα). At what rate does the biomass increase?

Detailed Solution:

The area of the colony is A = πr2 and dr/dt = C.

(a) dA/dt = π(2r)dr/dt = 2πrC.

(b) M = αA so dM/dt = αdA/dt = α2πrC.

7.4 Limb development

During early development, the limb of a fetus increases in size, but has a constant proportion.Suppose that the limb is roughly a circular cylinder with radius r and length l in proportion

l/r = C

where C is a positive constant. It is noted that during the initial phase of growth, the radiusincreases at an approximately constant rate, i.e. that

dr/dt = a.

At what rate does the mass of the limb change during this time? [Note: assume that the density ofthe limb is 1 gm/cm3 and recall that the volume of a cylinder is

V = Al

where A is the base area (in this case of a circle) and l is length.]

Detailed Solution:

We have the volume of the cylinderV = πr2l

andl = Cr

so thatV = πr2(Cr) = Cπr3.

The mass of the limb is volume times density so M = 1V = Cπr3. Thus

dM

dt= Cπ(3r2)

dr

dt= Cπ(3r2)a.

v.2005.1 - September 4, 2009 3

Page 4: chainRule_implicit differentiation

Math 102 Problems Chapter 7

7.5

A rectangular trough is 2 meter long, 0.5 meter across the top and 1 meter deep. At what ratemust water be poured into the trough such that the depth of the water is increasing at 1 m/minwhen the depth of the water is 0.7 m?

Detailed Solution:

See Figure 7.1. Let x and y be the dimensions of the base of the trough, and z be the depth of thewater in the trough. Then the volume of water in the trough V = xyz. x, y are constants and z is

a function of time t.dz

dt= 1.

dV

dt= xy

dz

dt= (2)(0.5)(1) = 1 m3/min

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��

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� � � �� � � �� � � �� � � �� � � �

z

y

x

Figure 7.1: Problem 7.5

7.6

Gas is being pumped into a spherical balloon at the rate of 3 cm3/s.

(a) How fast is the radius increasing when the radius is 15 cm?

(b) Without using the result from (a), find the rate at which the surface area of the balloon isincreasing when the radius is 15 cm.

Detailed Solution:

At time t the sphere has radius r, volume V = 4

3πr3, and surface area S = 4πr2.

v.2005.1 - September 4, 2009 4

Page 5: chainRule_implicit differentiation

Math 102 Problems Chapter 7

(a)

V =4

3πr3

dV

dt= 4πr2 ·

dr

dtdr

dt=

dVdt

4πr2

=3

4π · (15)2

=1

300πcm/s

(b)

S = 4πr2

dS

dt= 8πr ·

dr

dtdS

dt=

2

r·dV

dtdS

dt=

2

15· (3)

=2

5cm2/s

7.7

A point moves along the parabola y =1

4x2 in such a way that at x = 2 the x-coordinate is increasing

at the rate of 5 cm/s. Find the rate of change of y at this instant.

Detailed Solution:

y =1

4x2

dy

dt=

1

2x ·

dx

dt=

1

2(2) (5) = 5 cm/s

7.8 Boyle’s Law

In chemistry, Boyle’s law describes the behaviour of an ideal gas: This law relates the volumeoccupied by the gas to the temperature and the pressure as follows:

PV = nRT

where n, R are positive constants.

(a) Suppose that the pressure is kept fixed, by allowing the gas to expand as the temperature isincreased. Relate the rate of change of volume to the rate of change of temperature.

v.2005.1 - September 4, 2009 5

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Math 102 Problems Chapter 7

(b) Suppose that the temperature is held fixed and the pressure is decreased gradually. Relatethe rate of change of the volume to the rate of change of pressure.

Detailed Solution:

(a) If P is constant, we can rewrite the relationship in the form

V (t) =nR

PT (t) = kT (t)

where k = nR/P is a constant. Then

dV

dt= k

dT

dt=

nR

P

dT

dt.

(b) In this case T is constant, so we have

V (t) =nRT

P (t)=

C

P (t).

where C = nRT is a constant. Then by the reciprocal law,

dV

dt= −

C

P 2

dP

dt= −

nRT

P 2

dP

dt

7.9 Spread of a population

In 1905 a Bohemian farmer accidentally allowed several muskrats to escape an enclosure. Theirpopulation grew and spread, occupying increasingly larger areas throughout Europe. In a classicalpaper in ecology, it was shown by the scientist Skellam (1951) that the square root of the occupiedarea increased at a constant rate, k. Determine the rate of change of the distance (from the site ofrelease) that the muskrats had spread. For simplicity, you may assume that the expanding area ofoccupation is circular.

Detailed Solution:

According to Skellam, if we putf(t) = A(t)1/2,

where A(t) is the occupied area, thenf ′(t) = k.

If we make the simplifying assumption that the areas of expansion are circular, we can use the factthat the area of a circle, where both A(t) and r(t) are functions of time t, is

A(t) = πr(t)2

which is the same as saying that

f(t) = A(t)1/2 = π1/2r(t)

v.2005.1 - September 4, 2009 6

Page 7: chainRule_implicit differentiation

Math 102 Problems Chapter 7

Differentiating both sides with respect to t gives

f ′(t) = π1/2r′(t),

which we can rearrange to getr′(t) = π−1/2f ′(t) = π−1/2k

since we assumed that f ′(t) = k. The distance is thus increasing at a constant rate of π−1/2k.

7.10

A spherical piece of ice melts so that its surface area decreases at a rate of 1 cm2/min. Find therate that the diameter decreases when the diameter is 5 cm.

Detailed Solution:

Let r = r(t) be the radius at time t and D(t) = 2r(t) the diameter. Then the surface area is

S = 4πr2 = πD2

and sodS

dt= 2πD

dD

dt.

But dSdt

= −1 and thereforedD

dt= −

1

2πD.

When D = 5 we havedD

dt= −

1

10π.

7.11 A Convex lens

A particular convex lens has a focal length of f = 10 cm. The distance p between an object andthe lens, the distance q between its image and the lens and the focal length f are related by theequation:

1

f=

1

p+

1

q.

If an object is 30 cm away from the lens and moving away at 4 cm/sec, how fast is its image movingand in which direction?

Detailed Solution:

We can use the relationship

1

10=

1

p+

1

q

to solve for q:1

q=

1

10−

1

p=

p − 10

10p

v.2005.1 - September 4, 2009 7

Page 8: chainRule_implicit differentiation

Math 102 Problems Chapter 7

q =10p

p − 10.

(The focal length of the lens is constant, but the other two quantities are changing and depend ont.) Differentiating each side with respect to t:

dq

dt=

10p′(t)(p − 10) − p′(t)(10p)

(p − 10)2= −

100p′(t)

(p − 10)2.

We now use the fact that p(t) = 30, p′(t) = −4, at the given instant. Then

dq

dt= 400/202 = 1.

Therefore the image is moving toward the lens at 1 cm/sec.

7.12 A conical cup

Water is leaking out of a small hole at the tip of a conical paper cup at the rate of 1 cm3/min. Thecup has height 8 cm and radius 6 cm, and is initially full up to the top. Find the rate of change ofthe height of water in the cup when the cup just begins to leak. [Remark: the volume of a cone isV = (π/3)r2h.]

Detailed Solution:

See Figure 7.2. Let r be the radius of the top surface of the water and h the height of the water inthe conical cup at a given time. Then the volume of water is given by

V = (π/3)r2h

and we are given thatdV

dt= −1

cm3

min.

We want to find dhdt

when r = 6 cm and h = 8 cm. First we need an expression that relates r to h.In the cone, the proportions of base radius to height are 6 to 8. The water always forms a conicalshape and so, by similar triangles,

r

h=

6

8⇒ r =

6h

8.

Substituting this into the volume equation, we get:

V =π

3

36h2

64h =

π3h3

16.

To solve the problem, we now differentiate the volume:

dV

dt=

16

d(h3)

dt=

163h2

dh

dt.

and isolate dhdt

:dh

dt=

16

1

h2

dV

dt.

Substituting in our values for dVdt

and h gives:

dh

dt=

16

576π(−1) =

−1

36π≈ 0.00884 cm/min.

v.2005.1 - September 4, 2009 8

Page 9: chainRule_implicit differentiation

Math 102 Problems Chapter 7

r

6

8

water h

Figure 7.2: Conical Cup for problem 7.12 solution

7.13 Conical tank

Water is leaking out of the bottom of an inverted conical tank at the rate of1

10m3/min, and at

the same time is being pumped in the top at a constant rate of k m3/min. The tank has height6 m and the radius at the top is 2 m. Determine the constant k if the water level is rising at the

rate of1

5m/min when the height of the water is 2 m. Recall that the volume of a cone of radius r

and height h is

V =1

3πr2h.

Detailed Solution:

If V (t) is the volume of water in the tank at time t, then

dV

dt= k −

1

10.

But we also know that (by similar triangles) h/r = 6/2 = 3 so r = h/3 and

V =1

3πr2h =

πh3

27,

where r is the radius of the water and h is the height. Therefore

dV

dt=

πh2

dh

dt=

45

when h = 2 and dhdt

= 1

5. Plugging this into our original expression relating dV

dtand k, it follows that

k =1

10+

45≈ 0.379.

7.14 The gravel pile

Gravel is being dumped from a conveyor belt at the rate of 30 ft3/min in such a way that thegravel forms a conical pile whose base diameter and height are always equal. How fast is the heightof the pile increasing when the height is 10 ft? (Hint: the volume of a cone of radius r and height

h is V =1

3πr2h.)

v.2005.1 - September 4, 2009 9

Page 10: chainRule_implicit differentiation

Math 102 Problems Chapter 7

Detailed Solution:

Let V (t) denote the volume of gravel at time t. Then

V =1

3πr2h =

πh3

12

since d = h, or simply r = h/2. Therefore

dV

dt=

πh2

4

dh

dt.

But we know thatdV

dt= 30,

and therefore when h = 10 we get

dh

dt=

4

πh2

dV

dt=

120

100π=

6

5π≈ 0.38 ft/min.

7.15 The sand pile

Sand is piled onto a conical pile at the rate of 10m3/min. The sand keeps spilling to the base of thecone so that the shape always has the same proportions: that is, the height of the cone is equal tothe radius of the base. Find the rate at which the height of the sandpile increases when the heightis 5 m. Note: The volume of a cone with height h and radius r is

V =π

3r2h.

Detailed Solution:

We would like to find h′(t) at that point in time at which h(t) = 5m. (h′(t) will be in m/min). Weknow that V ′(t) = 10 for all t, and that

V (t) =π

3r(t)2h(t) =

π

3h(t)3,

since r(t) = h(t) for all t. Using the chain rule, we get

V ′(t) = πh(t)2h′(t)

and solving for h′(t) we find

h′(t) =V ′(t)

πh(t)2=

10

πh(t)2.

At the point in time when h(t) = 5, we therefore have

h′(5) =10

π25=

2

5π≈ 0.1273 m/min.

v.2005.1 - September 4, 2009 10

Page 11: chainRule_implicit differentiation

Math 102 Problems Chapter 7

7.16

Water is flowing into a conical reservoir at a rate of 4 m3/min. The reservoir is 3 m in radius and12 m deep.

(a) How fast is the radius of the water surface increasing when the depth of the water is 8 m?

(b) In (a), how fast is the surface rising?

Detailed Solution:

(a) V = 1

3πr2h. See Figure 7.3, the radius of the water surface is related to the water depth by

r

h=

3

12⇒ h = 4r. So V =

1

3πr2(4r) =

4

3πr3. When h = 8, r = 2.

12 m

h

3 m3 m

h

12 mr

Figure 7.3: Problem 7.16

dV

dt= 4πr2

dr

dtdr

dt=

dVdt

4πr2

=4

4π(2)2

=1

4πm/min

(b) h = 4r, sodh

dt= 4

dr

dt.

dh

dt= 4

1

4π=

1

πm/min.

7.17

A ladder 10 meters long leans against a vertical wall. The foot of the ladder starts to slide awayfrom the wall at a rate of 3 m/s.

(a) Find the rate at which the top of the ladder is moving downward when its foot is 8 metersaway from the wall.

v.2005.1 - September 4, 2009 11

Page 12: chainRule_implicit differentiation

Math 102 Problems Chapter 7

(b) In (a), find the rate of change of the slope of the ladder.

Detailed Solution:

Let x be the distance between the foot of the ladder and the foot of the wall and y be the distancebetween the foot of the wall and the top of the ladder. See Figure 7.4.

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y

x

3 m/s

10 m

Figure 7.4: Problem 7.17

(a) x, y and the length of the ladder satisfies x2 +y2 = 102 = 100. When x = 8, y =√

100 − 82 =6. Differentiate both sides of x2 + y2 = 100 with respect to t:

2xdx

dt+ 2y

dy

dt= 0

dy

dt=

−2xdxdt

2y

=−xdx

dt

y

=−8(3)

6= −4 m/s

(b) Slope s =y

x. Rate of change of slope is

ds

dt.

dS

dt=

dydt

· x − y · dxdt

x2

=(−4)(8) − 6(3)

82

= −25

32per sec

v.2005.1 - September 4, 2009 12

Page 13: chainRule_implicit differentiation

Math 102 Problems Chapter 7

7.18 Sliding ladder

A ladder 5 m long rests against a vertical wall. If the bottom of the ladder slides away from thewall at the rate of 0.5 meter/min how fast is the top of the ladder sliding down the wall when thebase of the ladder is 1 m away from the wall ?

Detailed Solution:

This is similar to problem 7.18. We relate the length of the ladder with its distance along the walland the ground using Pythagoras: x2 + y2 = 25. We differentiate this equation with respect to t:

2xdx

dt+ 2y

dy

dt= 0 =⇒

dy

dt= −

x

y

dx

dt.

But we have dxdt

= 1/2 and x = 1. We can find y by plugging x = 1 into the equation x2 + y2 =

25 ⇒ 1 + y2 = 25 ⇒ y =√

24 = 2√

6. We substitute these values into dydt

= −xy

dxdt

to get:

dy/dt =−x

2y=

−1

4√

6.

7.19

Ecologists are often interested in the relationship between the area of a region (A) and the numberof different species S that can inhabit that region. Hopkins (1955) suggested a relationship of theform

S = a ln(1 + bA)

where a and b are positive constants. Find the rate of change of the number of species with respectto the area. Does this function have a maximum?

Detailed Solution:

Using the chain rule we find thatdS

dA=

ab

1 + bA.

This derivative is never zero, which means that there are no local maxima. The larger the area, thegreater the number of species that would inhabit it according to this formula.

7.20 The burning candle

A candle is placed a distance l1 from a thin block of wood of height H . The block is a distancel2 from a wall as shown in Figure 7.5. The candle burns down so that the height of the flame, h1

decreases at the rate of 3 cm/hr. Find the rate at which the length of the shadow y cast by theblock on the wall increases. (Note: your answer will be in terms of the constants l1 and l2. Remark:This is a challenging problem.)

v.2005.1 - September 4, 2009 13

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Math 102 Problems Chapter 7

l l

Hh h

1

1 1

y

2

Figure 7.5: Figure for Problem 7.20

Detailed Solution:

Let h1(t) be the height of the candle flame at time t, let y(t) be the length of the shadow at timet, and let L = l1 + l2. Then L is constant. Also we will define h2 = H − h1. Note that h1 = h1(t)is changing with time and so is h2.

Looking at the geometry shown in Figure 7.5 above the dotted line, we see similar triangleshaving side lengths L, (y − h1), and h2, l1, Thus, by similar triangles,

y − h1

L=

h2

l1=

H − h1

l1.

The height of the shadow is therefore

y(t) = L

(

H − h1(t)

l1+

h1(t)

L

)

Differentiating with respect to time:

dy

dt= L

(

−1

l1

dh1

dt+

1

L

dh1

dt

)

.

Simplifying leads tody

dt=

dh1

dtL

(

−1

l1+

1

L

)

.

Using dh1/dt = −3 leads to

dy

dt= −3(l1 + l2)

(

−1

l1+

1

(l1 + l2)

)

.

After simplifying we getdy

dt= 3

(

l1 + l2l1

− 1

)

.

dy

dt= 3

l2l1

v.2005.1 - September 4, 2009 14

Page 15: chainRule_implicit differentiation

Math 102 Problems Chapter 7

7.21

Use implicit differentiation to show that the derivative of the function

y = x1/3

isy′ = (1/3)x−2/3.

First write the relationship in the form y3 = x, and then find dy/dx.

Detailed Solution:

Rewrite the equation in the formy3 = x

and then differentiate both sides with respect to x :

3y2y′ = 1.

Solving for y′ we get

y′ =1

(3y2)=

1

(3x2/3)= (1/3)x−2/3.

7.22 Generalizing the Power Law

(a) Use implicit differentiation to calculate the derivative of the function

y = f(x) = xn/m

where m and n are integers. (Hint: rewrite the equation in the form ym = xn first.)

(b) Use your result to derive the formulas for the derivatives of the functions y =√

x and y =x−1/3.

Detailed Solution:

(a) Since y = f(x) = xn/m, we have ym = xn. So d(ym)/dx = d(xn)/dx = nxn−1. Using the chainrule,

dym

dx=

dym

dy

dy

dx= mym−1

dy

dx

Therefore, mym−1(dy/dx) = nxn−1 so that

dy

dx=

n

m

xn−1

ym−1=

n

mxn−1

y

ym.

Substituting the expression y = xn/m in this last result, we get

dy

dx=

n

m

xn−1xn/m

xn=

n

mxn/m−1

v.2005.1 - September 4, 2009 15

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Math 102 Problems Chapter 7

(b) We have

y =√

x = x1/2.

Using the results in

(a) we get

dy/dx = (1/2)x1/2−1 = (1/2)x−1/2 = 1/(2√

x.)

For y = x−1/3 we get

dy/dx = (−1/3)x−1/3−1 = (−1/3)x−4/3.

7.23

The equation of a circle with radius r and center at the origin is

x2 + y2 = r2

(a) Use implicit differentiation to find the slope of a tangent line to the circle at some point (x, y).

(b) Use this result to find the equations of the tangent lines of the circle at the points whose xcoordinate is x = r/

√3.

(c) Use the same result to show that the tangent line at any point on the circle is perpendicularto the radial line drawn from that point to the center of the circle

Note: Two lines are perpendicular if their slopes are negative reciprocals.

Detailed Solution:

(a) We are asked to find the slope of the tangent line to a circle with radius r and center at theorigin, whose equation is

x2 + y2 = r2.

Taking the derivative with respect to x on both sides we get:

d

dx(x2 + y2) =

d

dx(r2)

2x +dy2

dx= 0

2x + 2ydy

dx= 0

Thus we can isolate y and find that

dy

dx= −

2x

2y= −

x

y.

v.2005.1 - September 4, 2009 16

Page 17: chainRule_implicit differentiation

Math 102 Problems Chapter 7

(b) We now want to find the equations of the tangent lines of the circle at the points whose xcoordinate is x = r

1/3. The corresponding y coordinate is given by

r2/3 + y2 = r2

i.e. y = ±r√

2/3.

Therefore, one of the points is (r√

1/3, r√

2/3). By part (a) the slope of the tangent line at

the point (r√

1/3, r√

2/3) would be

(dy/dx) = −x/y = −(r√

1/3)/(r√

2/3) = −1/√

2.

We can now use the point-slope method of finding the equation of the tangent line:

y − r√

2/3

x − r√

1/3= −

1√2

After simplifying we find that

y = −1√2

(

x −r√3

)

+ r

√2√3

=1√2

(

−x +r√3

+2r√

3

)

=1√2(−x + r

√3)

We can similarly do the calculations for the tangent line that goes through the point (r√

1/3,−r√

2/3)

and the result is y = (1/√

2)(x − r√

3). (Note: this second result can also be obtained byconsidering the symmetry of the two cases about the x axis.)

(c) The slope of a line connecting the origin, (0, 0), and a point on the circle, (x, y) is m = y/x.The slope of the tangent line at the point (x, y) on the circle is, by our previous result, −x/y.Since these slopes are negative reciprocals, the two lines are perpendicular.

7.24

The equation of a circle with radius 5 and center at (1, 1) is

(x − 1)2 + (y − 1)2 = 25

(a) Find the slope of the tangent line to this curve at the point (4, 5).

(b) Find the equation of the tangent line.

Detailed Solution:

(a) Using implicit differentiation we have

2(x − 1) + 2(y − 1)dy/dx = 0 =⇒ dy/dx = −(x − 1)/(y − 1).

At (4, 5) this gives −3/4 for the slope.

Another way to compute the slope is to use the fact that for a circle the tangent line isperpendicular to the radial line (the line from the center (1, 1) to (4, 5)). This slope is (5 −1)/(4 − 1) = 4/3 and therefore the slope of the tangent line must be −3/4.

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(b) The equation of the tangent line isy − 5

x − 4=

−3

4,

or y = −(3/4)x + 8.

7.25 Tangent to a hyperbola

The curvex2 − y2 = 1

is a hyperbola. Use implicit differentiation to show that for large x and y values, the slope dy/dxof the curve is approximately 1.

Detailed Solution:

Differentiating both sides :

d

dx(x2 − y2) =

d

dx1 =⇒ 2x − 2y(dy/dx) = 0.

Thus dy/dx = x/y. But on the hyperbola, x2 − y2 = 1 so that

x = ±√

1 + y2.

When x and y are very large compared to 1, it is approximately true that x ≈ ±√

y2 = ±y. Whenboth x and y are positive, this means that x ≈ y. Thus the slope of the tangent line for largepositive values of x and y is

dy/dx = x/y ≈ x/x = 1.

7.26 An ellipse

Use implicit differentiation to find the points on the ellipse

x2

4+

y2

9= 1

at which the slope is -1/2.

Detailed Solution:

For the ellipse x2

a2 + y2

b2= 1, implicit differentiation gives

2x

a2+

2y

b2

dy

dx= 0

solving for the slope, we find thatdy

dx= −

xb2

ya2.

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For the above ellipse, a = 2, b = 3 so that

dy

dx= −

9x

4y.

Thus, the points at which this slope is - 1/2 satisfy −9x4y

= −1/2. From the equation of the ellipse

we have the relation y =

9 −9x2

4. Plugging this into −9x

4y= −1/2 gives us:

−9x

12√

1 − x2

4

=−1

2=⇒

−9

12√

1

x2 − 1

4

=−1

2

1

x2−

1

4=

3

2=⇒

1

x2=

9

4+

1

4=

10

4

x2 =4

10=⇒ x = ±

2√10

Then y =

9 −9

4

(

4

10

)

= ±9√10

.

However, there should only be 2 points on the ellipse where the slope is −1/2, namely in the1st and 3rd quadrants. To verify that these are the correct solutions, we plug ( 2√

10, 9√

10) and

(− 2√10

,− 9√10

) into the equation −9x4y

= −1/2, and get −1/2 = −1/2 in both cases. Therefore the

points on the ellipse at which the slope is −1/2 are ( 2√10

, 9√10

) and (− 2√10

,− 9√10

).

7.27 Motion of a cell

In the study of cell motility, biologists often investigate a type of cell called a keratocyte, anepidermal cell that is found in the scales of fish. This flat, elliptical cell crawls on a flat surface,and is known to be important in healing wounds. The 2D outline of the cell can be approximatedby the ellipse

x2/100 + y2/25 = 1

where x and y are distances in µ (Note: 1 micron is 10−6 meters). When the motion of the cellis filmed, it is seen that points on the “leading edge” (top arc of the ellipse) move in a directionperpendicular to the edge. Determine the direction of motion of the point (xp, yp) on the leadingedge.

Detailed Solution:

The ellipse isx2/100 + y2/25 = 1

so that its major and minor axes are a = 10, b = 5. The slope of the ellipse (i.e. of the edge of thecell) at a given point is

dy/dx = −xb2

ya2= −

25x

100y= −

x

4y

A direction perpendicular to this would be a line whose slope is the negative reciprocal of dy/dx,namely m = 4y

xand at the point (xp, yp) this direction would be m = 4yp

xp

.

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Math 102 Problems Chapter 7

7.28 The Folium of Descartes:

A famous curve (see Figure 7.6) that was studied historically by many mathematicians (includingDescartes) is

x3 + y3 = 3axy

a

-a

(1.5a,1.5a)

Figure 7.6: The Folium of Descartes in Problem 7.28

You may assume that a is a positive constant.

(a) Explain why this curve cannot be described by a function such as y = f(x) over the domain−∞ < x < ∞.

(b) Use implicit differentiation to find the slope of this curve at a point (x, y).

(c) Determine whether the curve has a horizontal tangent line anywhere, and if so, find the xcoordinate of the points at which this occurs.

(d) Does implicit differentiation allow you to find the slope of this curve at the point (0,0) ?

Detailed Solution:

The Folium of Descartes is the curvex3 + y3 = 3axy

where a is a positive constant.

(a) This curve cannot be described by a function such as y = f(x) over the domain −∞ < x < ∞because it is not possible to solve for y in the equation and to obtain a single valued functiony = f(x). Further, from the picture of the curve shown on the homework sheet, we see thatit cannot be represented by a function since several values of x correspond to more than onevalue of y, i.e. the curve fails the vertical line property.

(b) We can use implicit differentiation to find the slope of this curve at a point (x, y). Theequation of the curve is x3 + y3 = 3axy so

3x2 + 3y2dy

dx= 3ay + 3ax

dy

dx

3(y2 − ax)dy

dx= 3(ay − x2)

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Math 102 Problems Chapter 7

dy

dx=

(ay − x2)

(y2 − ax)

(c) Horizontal tangent(s) to the curve occur when dy/dx = 0, i.e. when

(ay − x2) = 0 or y = x2/a

and when the equation of the curve is also satisfied. Plugging y = x2/a into the equation ofthe curve, we find that

x3 +x6

a3= 3a

x3

a=⇒ x6 = 2x3a3

which means that either x = 0 (in which case y = 0 too) or else x = 21/3a.

(d) At the point (0, 0), our result by implicit differentiation is that the slope of the curve is 0/0which is undefined. Thus we cannot use this method. We also note from the sketch of thecurve that there are two branches which both go through the origin. Thus there is not oneunique tangent line at this point, and the derivative is not defined there. (Note: It is possibleto discuss the slope of the various branches that cross the origin, but for this we would haveto investigate the so-called parametric form of this curve, something that will be discussed inMath 200.)

7.29 Isotherms in the Van-der Waal’s equation:

In thermodynamics, the Van der Waal’s equation relates the mean pressure, p of a substance to itsmolar volume v at some temperature T as follows:

(p +a

v2)(v − b) = RT

where a, b, R are constants. Chemists are interested in the curves described by this equation whenthe temperature is held fixed. (These curves are called isotherms).

(a) Find the slope, dp/dv, of the isotherms at a given point (v, p).

(b) Determine where points occur on the isotherms at which the slope is horizontal.

Detailed Solution:

The isotherms in the Van-der Waal’s equation are given by

(p +a

v2)(v − b) = RT (∗)

where a, b, R are constants. In this case, since T is held fixed, it is also a constant.

(a) We can find the slope using implicit differentiation. Using the product rule and implicitdifferentiation we get

(dp

dv− 2

a

v3)(v − b) + (p +

a

v2)(1) = 0.

Thusdp

dv= (2

a

v3) − (p +

a

v2)/(v − b).

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Math 102 Problems Chapter 7

(b) dpdv

= 0 when

(2a

v3) − (p +

a

v2)/(v − b) =

2a(v − b) − (pv3 + av)

v3(v − b)= 0

i.e. when2a(v − b) − pv3 − av = 0

Solving for p we get

p =2a(v − b) − va

v3=

a(v − 2b)

v3.

To find points on the isotherms at which the slope is 0, we need to make sure that the values

of p, v satisfy the equation of the isotherms, (*). Therefore we plug p =a(v − 2b)

v3into (*):

av − 2ab + av

v3(v−b) = RT =⇒ (2av−2ab)(v−b) = RTv3 =⇒ 2av2−2avb−2avb+2ab2 = RTv3

This leads to a cubic equation in v, namely

RTv3 − 2av2 + 4ahv − 2ab2 = 0.

We will find out how to solve such equations using Newton’s method later on in this course.

7.30 The circle and parabola

A circle of radius 1 is made to fit inside the parabola y = x2 as shown in figure 7.7. Find thecoordinates of the center of this circle, i.e. find the value of the unknown constant c. [Hint: Set upconditions on the points of intersection of the circle and the parabola which are labeled (a, b) in thefigure. What must be true about the tangent lines at these points?]

x

y

(0,c)

(a,b)

Figure 7.7: Figure for Problem 7.30

Detailed Solution:

The equation of the circle is(y − c)2 + x2 = 1.

We now gather the needed conditions to solve for c.

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Condition 1: (a, b) is on the parabola so that b = a2.Condition 2: (a, b) is on the circle so that (b − c)2 = 1 − a2.Condition 3: The parabola and the circle share a tangent line at the point (a, b). This implies

that the slope computed from dy/dx = 2x = 2a at this point matches the slope computed for thetangent to the circle.

To compute that slope, observe that the slope of the radius vector joining (a, b) to (0, c) is

(c − b)/(0 − a) = −(c − b)/a.

The tangent is perpendicular to this radius vector, and thus has slope which is the negative recip-rocal, i.e. a/(c − b).

We now know that a/(c − b) = 2a (from equating slopes) which means that c − b = 1/2. Using(b − c)2 = 1 − a2 now gives us 1 − a2 = 1/4 which implies that a =

√3/2 and b = a2 = 3/4. Thus

c = b + 1/2 = 5/4. The coordinate of the center of the circle is (0, 5/4).

7.31

Consider the curve whose equation is

x3 + y3 + 2xy = 4, y = 1 when x = 1.

(a) Find the equation of the tangent line to the curve when x = 1.

(b) Find y′′ at x = 1.

(c) Is the graph of y = f(x) concave up or concave down near x = 1?

Hint: Differentiate the equation x3 + y3 + 2xy = 4 twice with respect to x.

Detailed Solution:

(a) Differentiating the equation x3 + y3 + 2xy = 4 with respect to x, and not forgetting that ydepends on x, gives 3x2 + 3y2y′ + 2y + 2xy′ = 0. Putting x = 1 and y = 1 and then solvingfor y′ we get 3 + 3y′ + 2 + 2y′ = 0 =⇒ y′ = −1. Thus the equation of the tangent line isy − 1 = −1(x − 1).

(b) Differentiating the equation 3x2 + 3y2y′ + 2y + 2xy′ = 0 with respect to x gives

6x + 6yy′ + 3y2y′′ + 2y′ + 2y′ + 2xy′′ = 0.

Now set x = 1, y = 1 and y′ = −1 to get 6 − 6 + 3y′′ − 2 − 2 + 2y′′ = 0 =⇒ y′′ =4

5.

(c) Since y′′(1) > 0 the graph is concave up near x = 1.

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