CHAPTER 4 SUPPLEMENTARY EXERCISES 1. Simplify the following expressions
5.3 IMPLICIT DIFFERENTIATION AND RELATED RATES
We have noted the usefulness of the chain rule for differentiating composite functions f(g(x)). Sometimes, as variables are related by a function, (as in the composite function f(g(x))), they may also be related by an equation as inf(x)g(y), say, whereyis a function of x.Implicit differentiationarises when variables are (or can be) related.
Using the chain rule, one can determinedy/dxby differentiatingimplicitly.
Example 5.3.1 Direct Differentiation and the Chain Rule
Determine the derivative of y= (5x2+3x+1)5directly and using implicit differentiation.
Solution:
Taking the derivative directly and using the power rule yields 5(5x2+3x+1)4 d
dx(5x2+3x+1) =5(5x2+3x+1)4(10x+3) Implicitly, replace(5x2+3x+1)by u so that y=u5. Using the chain rule,
dy dx = dy
du⋅ du
dx =5u4(10x+3) =5(5x2+3x+1)4(10x+3)
In general, a rule for using the power rule in an implicit differentiation is:
The Implicit Differentiation Rule
d
dx(yr) =ryr−1dy dx
Example 5.3.2 Implicit Differentiation
Use implicit differentiation to calculatedy
dxof x3+y4=5x+4.
Solution:
Each term on both sides of the equation is differentiated with respect toxto yield 3x2+4y3
(dy dx
)
=5.
Next, solve for dy
dx. Therefore,dy
dx = 5−3x2 4y3 .
Example 5.3.3 More on Implicit Differentiation
Use implicit differentiation to obtain dy
dx for 9x2+2y3=5y.
Solution:
Each term on both sides of the equation is differentiated to yield 18x+6y2dy
dx =5dy dx. Next, solve for dy
dx. Therefore,
18x=5dy
dx −6y2dy dx 18x= (5−6y2)dy
dx dy
dx= 18x 5−6y2
Note that the previous two examples were relatively simple in that the variables did not appear as products or quotients. The following examples illustrate more complex situations.
IMPLICIT DIFFERENTIATION AND RELATED RATES 149
Example 5.3.4 Implicit Differentiation and the Product Rule
Use implicit differentiation to calculatedy
dx of 5x2y7=8x2+7x+3.
Solution:
Each term on both sides of the equation is differentiated to yield 5x2
[ 7y6dy
dx ]
+y7[10x] =16x+7 Notedy
dxdoesn’t appear in the second term, y7[10x], as the derivative is with respect tox.
Next, solve for dy dx.
(35x2y6)dy
dx= −10xy7+16x+7 dy
dx= −10xy7+16x+7 35x2y6 Economics Application – Cobb – Douglas Function
Economists use a Cobb– Douglas Production Function to model, say, a manufacturing enterprise. It takes the form Ax𝛼y𝛽 =q, wherexandyare units of capital and labor, respec- tively; A is a constant, andqrepresents the quantity of goods produced. When𝛼+𝛽 =1, the function is said to exhibit “constant returns to scale”. This means that doubling inputs (xandy), doubles outputs (q); tripling inputs, triples outputs, and so on (also see page 221).
Example 5.3.5 Cobb–Douglas Production Function
Calculate dy
dxusing implicit differentiation of the production function 80x1∕3y2∕3=960.
Solution:
Differentiating implicitly and using the chain rule yield 80x1∕3
[2 3y−1∕3dy
dx ]
+y2∕3 [80
3 x−2∕3 ]
=0 Next, solve for dy
dx as 80x1∕3
[2 3y−1∕3dy
dx ]
= −y2∕3 [80
3 x−2∕3 ]
dy dx=
−y2∕3 [80
3 x−2∕3 ] 160
3 x1∕3y−1∕3
= −y 2x.
↑ The absolute value of dy
dx is called the marginal rate of substitution (MRS) of x for y.
Economists regard the MRS as the rate at which one is willing to substitute one good for another without loss or gain.
Related Rates
Many everyday quantities of interest change with time! Interesting situations arise when two or more rates of change are related to each other. Examples abound!
• The top of a ladder propped against a wall falls at one rate as the foot of the ladder moves away from the wall at a different rate.
• An airplane in flight moves at one speed while its ground shadow recedes at another rate.
• Sales of a product increase at one rate while the related advertising expenditure increases at still another rate.
• A chemist stirs at one rate while a substance dissolves in a solvent at another rate, and so on.
In each of these instances, a mathematical relation−sometimes geometric−connects the two rates. Differentiation with respect to time connects the related rates.
A series of examples helps you gain useful skills.
Example 5.3.6 Related Rates
A 25-foot ladder rests on a horizontal surface and leans against a vertical wall. If the bottom of the ladder slides from the wall at a rate of 3 ft/s, how fast is the ladder sliding down the wall when the top of the ladder is 15 ft from the ground?
Solution:
Let x be the ladder’s distance from the base of the wall and y its height on the wall. Note that both x and y are actually functions of time. The length of the ladder being fixed, the variables x and y are related geometrically by a right triangle as
x2+y2=252 =625
As we are interested in a time-related rate, we differentiate with respect to time, t, even as it does not appear explicitly.
2xdx dt +2ydy
dt =0 or equivalently, xdx
dt +ydy dt =0
IMPLICIT DIFFERENTIATION AND RELATED RATES 151 At an instant when y is 15 ft, x is 20 ft. At this instant, the rate at which the ladder moves from the wall is dx
dt =3 ft∕s. To solve fordy/dt, the rate at which it moves down the wall substitute,
(20)(3) + (15)dy dt =0 dy dt = −4
At the instant that x=20 ft and y=15 ft, the ladder is moving down the side of the wall at a rate of 4 ft/s.
Example 5.3.7 A Water Tank
A water tank in the shape of a right circular cone has radius 8 ft and height 16 ft. Water is pumped into the tank at a rate of 20 gallons per minute. Clearly, the rate at which water is pumped and its depth in the tank are related. At what rate is the water level rising when it is 2 ft deep(1 gal≈0.134 ft3)?
Solution:
Let r be the radius at the water surface when its depth is h. The volume of a right circular cone is V=1∕3 πr2h. Asrandhare functions of time, t, we seekdh
dt, the rate at which the water level is rising.
For this cone, r=h∕2 as the ratio of r tohis 8/16. When the depth ish, the volume is V= 1
3 π (h
2 )2
h= 1 12πh3.
Next, differentiation with respect to time, t, yieldsdV dt = 1
4πh2 (dh
dt )
. From the data, the change in volume,
dV
dt = (20 gals∕min)(0.134 ft3∕gal) =2.68 ft3∕min when h=2 ft.
SolvingdV dt for dh
dt yields dh
dt = 4 πh2
(dV dt )
= 4
π(2 ft)2(2.68 ft3∕min) =0.853 ft∕min
↓ To solve related rate problems, a geometric or other relation between the variables is necessary. Then obtain expressions for the derivatives at an arbitrary time and substitute for the instantaneous values of variables.
EXERCISES 5.3
In Exercises 1–14, suppose that x and y are related as indicated. Calculate dy/dx implic- itly.
1. 4x2+9y2=25 2. 2x3−y2=1 3. 2x3+7y4=3x+7 4. x3−2x2=y2+1 5. x2−2x+y3=3y+7 6. 1
x3 + 1 y2 =1 7. 1
x2 + 3 y4 =14x
8. xy=6 9. x2y2=18x 10. x2y+3x=6y 11. x3y5+2x=6x2 12. 3x2−2xy+4y3=x3 13. 4xy3−x2y+x3+3=9x 14. 5x−y=3y2+2y3+7x5
In Exercises 15–19, obtain the indicated slope by differentiating implicitly.
15. xy+15=0 at(−3,5) 16. x2+y2=169 at(12,5)
17. y4+2y−3x3=2x+8 at(−1,1) 18. √3
x+√
y=3 at(8,1) 19. xy+y3=18 at(5,2)
20. Determine the tangent tox4y2=4 at(1,2)and(1,−2) 21. Determine the tangent toxy4 =48 at(3,2)
In Exercises 22–25, x and y are differentiable functions of time, t, and are related by the given equation. Use implicit differentiation to find dy
dt in terms of x, y, and dx dt. 22. x2+y2=25
23. x4+y4=17 24. 5x2y−3y=2x+5 25. x2+4xy=7x+y2
26. A 13-foot ladder rests against a wall. If the bottom of the ladder slides horizontally from the wall at 2 ft/s, how fast is the top of the ladder sliding down the wall when it is 12 ft above the ground?
27. Air is pumped into a spherical balloon at a rate of 3 ft3∕min. Find the rate at which the radius is changing when its diameter is 12 inches, assuming constant pressure.
28. A kite is flying at a height of 60 ft. Wind carries the kite horizontally away from an observer at a rate of 2 ft/s. At what rate must the string be released when the kite is 100 ft away?
FINITE DIFFERENCES AND ANTIDIFFERENCES 153 29. The current,I, (in amperes) in an electrical circuit isI=100∕RwhereRis the resis- tance (in ohms). Find the rate of change ofIwith respect toRwhen the resistance is 5 ohms.
30. Suppose that the pricep(in dollars) and the monthly salesx(in thousands of units) of a commodity satisfy 3p3+2x2=3800. Determine the rate at which sales are changing at a time whenx=20,p=10 and the price is falling at a rate of $1 per month.
31. Consider the Cobb– Douglas Production Function, 20x1∕4y3∕4 =1080. Use implicit differentiation to calculate dy
dx whenx=16 andy=81.