I V=iirr\ So, D,V=4irr
2 -D,r. We are told that D,V= 2. So, 2 = 4irr
2 -D,r. When r=|, 2 =
4ir($)-D,r, D,r = 2lir. Let d be the diameter. Then, d = 2r, D,d = 2- D,r = 2- (21 IT) = 4/ir~ 1.27.
So, the diameter is increasing at the rate of about 1.27 inches per second.
14.8
Oil from an uncapped well in the ocean is radiating outward in the form of a circular film on the surface of the
water. If the radius of the circle is increasing at the rate of 2 meters per minute, how fast is the area of the oil film
growing when the radius is 100 meters.
I The area A = irr
2 . So, D,A = 2irr- D,r. We are given that D,r = 2. Hence, when r = 100,
D,A = 277 • 100 • 2 = 4007T, which is about 1256 m
2 /min.
14.9
The length of a rectangle of constant area 800 square millimeters is increasing at the rate of 4 millmeters per
second. What is the width of the rectangle at the moment the width is decreasing at the rate of 0.5 millimeter per
second?
I The area 800 = fw. Differentiating, 0= (• D,w + w D,f. We are given that D/ = 4. So, 0 =
f-D,w + 4w. When D,w=-0.5, 0=-0.5/ + 4w, 4w = 0.5^. But
2 = 100, w = 10 mm.
14.10
Under the same conditions as in Problem 14.9, how fast is the diagonal of the rectangle changing when the width is
20mm?
I As in the solution of Problem 14.9. 0 = (• D,w + 4w. Let u be the diagonal. Then u
2 = w
2 + f
2 ,
2u-D,u = 2wD,w + 2f-D,t, u • D,u = w D,w + (• D,f. When w = 20,
When w = 20, u
2 = (20)
2 + (40)
2 = 2000, u = 20V5.
Substituting in u • D,u = w • D,w + t- D,f, 20V5 • D,u = 20-(-2) + 40-4= 120, D,u = 6V5/5 = 2.69mm/s.
14.11
A particle moves on the hyperbola x
2 — I8y
2 = 9 in such a way that its ^-coordinate increases at a constant rate
of 9 units per second. How fast is its ^-coordinate changing when x = 91
I 2x-D,x-36yD t y = 0, x • D,x = l8y • D,y. We are given that D,y = 9. Hence, x- D,x = 18y -9 =
I62y. When * = 9, (9)
2 - 18y
2 = 9, I8y
2 = 72, y
2 =4, y = ±2. Substituting in x • D,x = 162y,
9 • D,x = ±324, D,x = ±36 units per second.
14.12
An object moves along the graph of y — f(x). At a certain point, the slope of the curve is | and the ^-coordinate
of the object is decreasing at the rate of 3 units per second. At that point, how fast is the y-coordinate of the
object changing?
I y=f(x). By the chain rule, D,y = /'(*) • D,x. Since f'(x) is the slope \ and D,x=-3, D,y =
j • (-3) = — 1 units per second.
14.13 If the radius of a sphere is increasing at the constant rate of 3 millimeters per second, how fast is the volume
changing when the surface area 4irr
2 is 10 square millimeters?
I y=|7rr
3 . Hence, D,V=4-!rr
2 • D,r. We are given that D,r = 3. So, D,V=47r/2 -3 When
47rr
2 = 10, D,y=30mm
3 /s.
14.14 What is the radius of an expanding circle at a moment when the rate of change of its area is numerically twice as
large as the rate of change of its radius?
I A = Trr
2 . Hence, D,A = 2irr • D,r. When D,A = 2-D,r, 2- D,r = 2irr- D,r, 1 = irr, r=\lir.
14.15
A particle moves along the curve y = 2jc
3 - 3x
2 + 4. At a certain moment, when x = 2, the particle's
^-coordinate is increasing at the rate of 0.5 unit per second. How fast is its y-coordinate changing at that
moment?
I D l y = 6x
2 -D t x-6x-D,x = 6x-D,x(x- 1). When x = 2, D,* = 0.5. So, at that moment, D, y =
12(0.5)(1) = 6 units per second.
14.16 A plane flying parallel to the ground at a height of 4 kilometers passes over a radar station R (Fig. 14-5). A short
time later, the radar equipment reveals that the distance between the plane and the station is 5 kilometers and that
the distance between the plane and the station is increasing at a rate of 300 kilometers per hour. At that moment,
how fast is the plane moving horizontally?
90
CHAPTER 14
2 -D,r. We are told that D,V= 2. So, 2 = 4irr
2 -D,r. When r=|, 2 =
4ir($)-D,r, D,r = 2lir. Let d be the diameter. Then, d = 2r, D,d = 2- D,r = 2- (21 IT) = 4/ir~ 1.27.
So, the diameter is increasing at the rate of about 1.27 inches per second.
14.8
Oil from an uncapped well in the ocean is radiating outward in the form of a circular film on the surface of the
water. If the radius of the circle is increasing at the rate of 2 meters per minute, how fast is the area of the oil film
growing when the radius is 100 meters.
I The area A = irr
2 . So, D,A = 2irr- D,r. We are given that D,r = 2. Hence, when r = 100,
D,A = 277 • 100 • 2 = 4007T, which is about 1256 m
2 /min.
14.9
The length of a rectangle of constant area 800 square millimeters is increasing at the rate of 4 millmeters per
second. What is the width of the rectangle at the moment the width is decreasing at the rate of 0.5 millimeter per
second?
I The area 800 = fw. Differentiating, 0= (• D,w + w D,f. We are given that D/ = 4. So, 0 =
f-D,w + 4w. When D,w=-0.5, 0=-0.5/ + 4w, 4w = 0.5^. But
2 = 100, w = 10 mm.
14.10
Under the same conditions as in Problem 14.9, how fast is the diagonal of the rectangle changing when the width is
20mm?
I As in the solution of Problem 14.9. 0 = (• D,w + 4w. Let u be the diagonal. Then u
2 = w
2 + f
2 ,
2u-D,u = 2wD,w + 2f-D,t, u • D,u = w D,w + (• D,f. When w = 20,
When w = 20, u
2 = (20)
2 + (40)
2 = 2000, u = 20V5.
Substituting in u • D,u = w • D,w + t- D,f, 20V5 • D,u = 20-(-2) + 40-4= 120, D,u = 6V5/5 = 2.69mm/s.
14.11
A particle moves on the hyperbola x
2 — I8y
2 = 9 in such a way that its ^-coordinate increases at a constant rate
of 9 units per second. How fast is its ^-coordinate changing when x = 91
I 2x-D,x-36yD t y = 0, x • D,x = l8y • D,y. We are given that D,y = 9. Hence, x- D,x = 18y -9 =
I62y. When * = 9, (9)
2 - 18y
2 = 9, I8y
2 = 72, y
2 =4, y = ±2. Substituting in x • D,x = 162y,
9 • D,x = ±324, D,x = ±36 units per second.
14.12
An object moves along the graph of y — f(x). At a certain point, the slope of the curve is | and the ^-coordinate
of the object is decreasing at the rate of 3 units per second. At that point, how fast is the y-coordinate of the
object changing?
I y=f(x). By the chain rule, D,y = /'(*) • D,x. Since f'(x) is the slope \ and D,x=-3, D,y =
j • (-3) = — 1 units per second.
14.13 If the radius of a sphere is increasing at the constant rate of 3 millimeters per second, how fast is the volume
changing when the surface area 4irr
2 is 10 square millimeters?
I y=|7rr
3 . Hence, D,V=4-!rr
2 • D,r. We are given that D,r = 3. So, D,V=47r/2 -3 When
47rr
2 = 10, D,y=30mm
3 /s.
14.14 What is the radius of an expanding circle at a moment when the rate of change of its area is numerically twice as
large as the rate of change of its radius?
I A = Trr
2 . Hence, D,A = 2irr • D,r. When D,A = 2-D,r, 2- D,r = 2irr- D,r, 1 = irr, r=\lir.
14.15
A particle moves along the curve y = 2jc
3 - 3x
2 + 4. At a certain moment, when x = 2, the particle's
^-coordinate is increasing at the rate of 0.5 unit per second. How fast is its y-coordinate changing at that
moment?
I D l y = 6x
2 -D t x-6x-D,x = 6x-D,x(x- 1). When x = 2, D,* = 0.5. So, at that moment, D, y =
12(0.5)(1) = 6 units per second.
14.16 A plane flying parallel to the ground at a height of 4 kilometers passes over a radar station R (Fig. 14-5). A short
time later, the radar equipment reveals that the distance between the plane and the station is 5 kilometers and that
the distance between the plane and the station is increasing at a rate of 300 kilometers per hour. At that moment,
how fast is the plane moving horizontally?
90
CHAPTER 14