# Inter 1st Year Maths 1B Applications of Derivatives Solutions Ex 10(e)

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(e) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(e)

I.

Question 1.
At time t, the distance s of a particle moving in a straight line is given by s = -4t² + 2t. Find the average velocity between t = 2 sec and t = 8 sec.
Solution:
s = -4t² + 2t ds
v = $$\frac{ds}{dt}$$ = -8t + 2 dt
Velocity at t = 2 is v = ($$\frac{ds}{dt}$$)t=2
v = -16 + 2 = -14 units/sec.
Velocity at t = 8 is v = ($$\frac{ds}{dt}$$)t=8
v = -64 + 2 = -62
Average velocity = $$\frac{-62-14}{2}$$ = -38 units/sec.

Question 2.
If y = x4 then find the average rate of change of y between x = 2 and x = 4.
Solution:
y = x4 ⇒ $$\frac{dy}{dt}$$ = 4x³
($$\frac{dy}{dt}$$)x=2 = 32
($$\frac{dy}{dt}$$)x=4 = 256
Average rate of change = $$\frac{256+32}{2}$$ = 144.

Question 3.
A particle moving along a straight line has the relation s = t³ + 2t + 3, connecting the distance s describe by the particle in time t. Find the velocity and acceleration of the particle of t = 4 sec.
Solution:
s = t³ + 2t + 3
$$\frac{ds}{dt}$$ = 3t² + 2, velocity v = $$\frac{ds}{dt}$$ = 3t² + 2
Velocity at t = 4
⇒ ($$\frac{ds}{dt}$$)t=4 = 48 + 2 = 50 units/sec
v = 3t² + 2
$$\frac{dv}{dt}$$ = 6t ⇒ a = ($$\frac{dv}{dt}$$)t=4 = 24 units/sec². Question 4.
The distance – time formula for the motion of a particle along a straight line is s = t³ – 9t² + 24t – 18. Find when and where the velocity is zero.
Solution:
Given s = t³ – 9t² + 24t – 18
v = $$\frac{ds}{dt}$$ = 3t² – 18t + 24
v = 0 ⇒ 3(t² – 6t + 8) = 0
∴ (t – 2) (t – 4) = 0
∴ t = 2 or 4
The velocity is zero after 2 and 4 seconds.

Case (i):
t = 2
s = t³ – 9t² + 24t – 18
= 8 – 36 + 48 – 18 = 56 – 54 = 2

Case (ii) :
t = 4 ; s = t³ – 9t² + 24t – 18
= 64 – 144 + 96 – 18
= 160 – 162 = -2
The particle is at a distance of 2 units from the starting point ‘O’ on either side.

Question 5.
The displacement s of a particle travelling in a straight line in t seconds is given by s = 45t + 11t² – t³. Find the time when the particle comes to rest.
Solution:
s = 45t + 11t² – t³
v = $$\frac{ds}{dt}$$ = 45 + 22t – 3t²
If a particle becomes to rest
⇒ v = 0 ⇒ 45 + 22t – 3t² = 0
⇒ 3t² – 22t – 45 = 0
⇒ 3t² – 27t + 5t – 45 = 0
⇒ (3t + 5) (t – 9) = 0
∴ t = 9 or t = – $$\frac{5}{3}$$
∴ t = 9
∴ The particle becomes to rest at t = 9 seconds.

II.

Question 1.
The volume of a cube is increasing at the rate of 8 cm³/sec. How fast is the surface area increasing when the length of an edge is 12 cm?
Solution:
Suppose ‘a’ is the edge of the cube and v be the volume of the cube.
v = a³ ……………….. (1)
$$\frac{dv}{dt}$$ = 8 cm³/sec.
a = 12 cm
Surface Area of cube. S = 6a²
$$\frac{ds}{dt}$$ = 12a$$\frac{da}{dt}$$ ………….. (2) Question 2.
A stone is dropped into a quiet lake and ripples move in circles at the speed of 5 cm/sec. At the instant when the radius of circular ripple is 8 cm., how fast is the enclosed area increases?
Solution:
Suppose r is the value of the outer ripple and A be its area
Area of circle A = πr²
$$\frac{dA}{dt}$$ = 2πr $$\frac{dr}{dt}$$
Given r = 8, $$\frac{dr}{dt}$$ = 5
$$\frac{dA}{dt}$$ = 2π (8) (5)
= 80π cm²/sec. Question 3.
The radius of a circle is increasing at the rate of 0.7 cm/sec. What is the rate of increase of its circumference?
Solution:
$$\frac{dr}{dt}$$ = 0.7 cm/sec
Circumference of a circle, c = 2πr
$$\frac{dc}{dt}$$ = 2π $$\frac{dr}{dt}$$
= 2π(0.7)
= 1.4π cm/sec.

Question 4.
A balloon which always remains spherical on inflation is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of balloon increases when the radius in 15 cm.
Solution:
$$\frac{dv}{dt}$$ = 900 c.c/sec
r = 15 cm
Volume of the sphere v = $$\frac{4}{3}$$ πr³ Question 5.
The radius of an air bubble is increasing at the rate of $$\frac{1}{2}$$ cm/sec. At what rate is the volume of the bubble increasing when the radius is 1 cm.?
Solution:
$$\frac{dr}{dt}=\frac{1}{2}$$ cm/sec
radius r = 1 cm
Volume sphere v = $$\frac{4}{3}$$ πr³
$$\frac{dv}{dt}$$ = 4πr² $$\frac{dr}{dt}$$
= 4π(1)²$$\frac{1}{2}$$
= 2π cm³/sec.

Question 6.
Assume that an object is launched upward at 980 m/sec. Its position would be given by s = 4.9 t² + 980 t. Find the maximum height attained by the object.
Solution:
s = – 4.9 t² + 980 t
$$\frac{ds}{dt}$$ = -9.8 t + 980
v = -9.8 t + 980
for max. height, v = 0
-9.8 t + 980 = 0
980 = 9.8 t
$$\frac{980}{9.8}$$ = t
100 = t
s = -4.9(100)² +980(100)
s = -49000 + 98000
s = 49000 units.

Question 7.
Let a kind of bacteria grow in such a way that at time t sec. there are t(3/2) bacteria. Find the rate of growth at time t = 4 hours.
Solution:
Let g be the amount of growth of bacteria at t then g(t) = t3/2
The growth rate at time t is given by
g'(t) = $$\frac{3}{2}$$t1/2
given t = 4hr
g'(t) = $$\frac{3}{2}$$ (4 × 60 × 60)1/2
= $$\frac{3}{2}$$ (2 × 60) = 180 Question 8.
Suppose we have a rectangular aquarium with dimensions of length 8m, width 4 m and height 3 m. Suppose we are tilling the tank with water at the rate of 0.4 m³/sec. How fast is the height of water changing when the water level is 2.5 m?
Solution:
Length of aquarium l = 8 m
Width of aquarium b = 4 m
Height of aquarium h = 3
$$\frac{dv}{dt}$$ = 0.4 m³/sec.
v = lbh
= 8(4)(3)
= 96
v = lbh
⇒ log v = log l + log b + log h Note : Text book Ans. $$\frac{1}{80}$$ will get when h = 3.

Question 9.
A container is in the shape of an inverted cone has height 8m and radius 6m at the top. If it is filled with water at the rate of 2m³/minute, how fast is the height of water changing when the level is 4m?
Solution: h = 8m = OC
r = 6m = AB
$$\frac{dv}{dt}$$ = 2 m³/minute
∆ OAB and OCD are similar angle then
$$\frac{CD}{AB}=\frac{OC}{OA}$$
$$\frac{r}{6}=\frac{h}{8}$$
r = h $$\frac{3}{4}$$
Volume of cone v = $$\frac{1}{3}$$πr²h Question 10.
The total cost C(x) in rupees associated with the production of x units of an item is given by C(x) 0.007x³ – 0.003x² + 15x + 4000. Find the marginal cost when 17 units are produced.
Solution:
Let m represents the marginal cost, then
M = $$\frac{dc}{dx}$$
Hence
M = $$\frac{d}{dx}$$(0.007x³ – 0.003x² + 15x + 4000)
= (0.007) (3x²) – (0.003) (2x) + 15 /.
∴ The marginal cost at x = 17 is
(M)m=17 = (0.007) 867 – (0.003) (34) + 15
= 6.069-0.102+ 15
= 20.967.

Question 11.
The total revenue in rupees received from the sale of x units of a produce is given by R(x) = 13x² + 26x + 15. Find the marginal revenue when x = 7.
Solution:
Let m denotes the marginal revenue. Then
M = $$\frac{dR}{dx}$$
Similar R(x) = 13x² + 26x +15
∴ m = 26x + 26
The marginal revenue at x = 7
(M)x = 7 = 26(7) + 26
= 208. Question 12.
A point P is moving on the curve y = 2x². The x co-ordinate of P is increasing at the rate of 4 units per second. Find the rate at which y co-ordinate is increasing when the point is (2, 8).
Solution:
Given y = 2x²
$$\frac{dy}{dx}$$ = 4x. $$\frac{dx}{dt}$$
Given x = 2, $$\frac{dx}{dt}$$ = 4.$$\frac{dy}{dt}$$
= 4(2).4 = 32
y co-ordinate is increasing at the rate of 32 units/sec.