calculus and analytical geometry lecture # 9 mth 104
TRANSCRIPT
Calculus and Analytical Geometry
Lecture # 9
MTH 104
Chain rule
1002 1xdxd
1 and 1let 21002 xuxy100then uy
Consider
xdxdu
ududy
2 and 100 99
dxdu
dudy
dxdy
)1(200
2)1(100
2100
2
2
99
xx
xx
xu
multiply rates
Chain rule
)( and ))(( xguxgfy
dxdu
dudy
dxdy
)cos( if Find 3xydxdy
If g is differentiable at x and f is differentiable at g(x) then the composition fog is differentiable at x. Moreover, if
Then y=f(u) and
Example
Alternatively
xgxgfxgfxgfdxd
Derivative of outside function Derivative of
inside function
Let 3xu cosuy Then
dudy usin
And
dxdu 23x
dxdu
dudy
dxdy
)3()sin( 2xu
32 sin3 xx
Rates of change multiply
Example
xdxd
2tan 2
tan xdxd x
dxd
x tan)tan2( xx 2sectan2
12xdxd 2
12 1x
dxd 11
21
22
12
xdxd
x
12
121
22
xx
xx
Derivative of outside function Derivative of
inside function
More examples
2 71) ( ) (3 5 )f x x x 297
5)()2
ttf
Solution 1)
xdxdu
xxu 103253Let
677Then ududy
uy
dxdu
dudy
dxdy
)103(67 xudxdy
)103(6)253(7 xxx
297
5)()2
ttf
97Let tu 7dtdu
2525 uu
y 310 ududy
dtdu
dudy
dtdy
7310 udxdy
3)97(70 tdtdy
Generalized derivative formulas
)( re whe)()( xgudxdu
ufufdxd
dxduuuu
dxd
dxduuuu
dxd
dxduuu
dxd
dxduuu
dxd
dxduuu
dxd
dxduuu
dxd
dxdurruru
dxd
cotcsccsc
tansecsec 2csccot
2sectan sincos
cossin 1
Some examples are:
Example
sin(2 )d
xdx
cos 2 2d
x xdx
2cos 2x
13
3
12 3
2 3d d
x xdx x x dx
2
3 32 3 2 3d
x x x xdx
23 22 3 3 2x x x
2
23
3 22 3x
x x
1.
2.
Example Find
2
2
2 if sin(3 )
d yy x
dx
Solution 2sin(3 )
dy dx
dx dx
2cos(3 )x 23d
xdx
26 cos(3 )x x
2
2
26 cos(3 )
d y dx x
dx dx
2 26 cos 3 cos(3 ) 6d d
x x x xdx dx
2 26 cos 3 cos(3 ) 6d d
x x x xdx dx
2 26 sin(3 ) 3d
x x xdx
26cos(3 )x
2 26 6 sin(3 ) 6cos 3x x x x
2
2 2 2
236 sin 3 6cos 3
d yx x x
dx
Example Differentiate 1ln 2 xy
1ln 2 xy
)2(1
12
xx
y
1
22
x
xy
Use the Chain Rule
Example Differentiatex
xy
ln
x
xy
ln Use the Quotient
Rule
2ln
1)1(ln
x
xxx
y
2ln
1ln
x
xy
Related rates
Consider a water is draining out of a conical filter. The volume V, the height h and the radius r are all functions of the elapsed time t.
hrV 2
3
rate of change of dV
Vdt
2[ ]3
dV dr h
dt dt
2[2 ]3
dV dr dhhr r
dt dt dt
The rate of change of V is related to the rates of change both r and h
Ralated rates
problem
Volume formula:
Example Suppose that x and y are differentiable functions of t and are related by the equation . Find dy/dt at time t=1 if x=2 and dx/dt=4 at time t=1.
3y x
solution
3y xDifferentiating both sides with respect to t
3dy dx
dt dt
23dx
xdt
2
1 1
3(2)
t t
dy dxdt dt
12 4 48
t=1 t=1
2, 4Known: Unknown : dx dydt dt
x
Example Suppose x and y are both differentiable functions of t and are related by the equation y = x2 + 3. Find dy/dt, given that dx/dt = 2 when x = 1
Solution
y = x2 + 3
Given dx/dt = 2 when x = 1
To find dy/dt
]3[][ 2 xdt
dy
dt
d
dt
dxx
dt
dy2
1 1
2(1) 2 2 4
x x
dy dxdt dt
Procedure for solving related rates problems
Step 1. Assign letters to all quantities that vary with time and any others that seem relevant to the problem. Give a definition for each letter.
Step 2. Identify the rates of change that are known and the rates of change that is to be found. Interpret each rate as a
derivative.Step 3. Find an equation that relates the variables whose rates of
change were identified in Step 2. To do this, it will often be helpful to draw an appropriately labeled figure that
illsutrates the relationship.Step 4. Differentiate both sides of the equation obtained in Step 3
with respect to time to produce a relationship between the known rates and the unknown rates of change.
Step 5. After completing Step 4, substitute all known values for the rates of change and the variables, and then solve for the unknown rate of change.
Example Suppose that , where both x and y are changing with time. At a certain instant when x=1 and y=2, x is decreasing at the rate of 2 units/s, and y is increasing at the rate of 3 units/s. How fast is z changing at this instant? Is z increasing or decreasing?
3 2z x y
solution
1 12 2
2, Known 3 :x xy y
dx dydt dt
12
Unknown: xy
dzdt
3 2z x y
Differentiating with respect to t
3 2dz dx y
dt dt
2 3 3 2dz d dy x x y
dt dt dt
2 2 33 2dx dy
x y x ydt dt
2 2 3
1 1 12 2 2
3(1) (2) 2(1) (2)x x xy y y
dz dx dydt dt dt
12
12( 2) 4(3) 12 units/sxy
dzdt
Negative sign shows that it is decreasing
Example A stone is dropped into a still pond sends out a circular ripple whose radius increases at a constant rate of 3ft/s. How rapidly is the area of enclosed by the ripple increases at the end of 10 s?
Solution
let r= radius of circular ripple, A= Area enclosed by the ripple
10
dr 3, To find
dtr
GivendAdt
Since radius is increasing with a constant rate of 3 ft/s, so after 10 s the radius will be 30 ft.
We know that 2A r
2dA dr
rdt dt
2
10
2 (30)(3) 180 /
r
ft sdAdt