Referring to the AP Inter 2nd Year Maths Study Material Chapter 6 Application of Derivatives Exercise 6a Solutions makes it easier to understand complex problems.
AP Inter 2nd Year Maths Application of Derivatives Solutions Exercise 6a
I.
Question 1.
Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cm
Solution:
(a) r = 3 cm
For the circle, we take radius = r and area = A
Given r = 3 cm, Area of circle A = πr2
Here, we have to find the rate of change of area A. w.r.t. ‘r’.
∴ A = πr2 ⇒ \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = π(2r) = 2πr
When r = 3,
\(\frac{\mathrm{dA}}{\mathrm{dr}}\) = 2π(3) = 6π
Thus, the area of the circle is changing at the rate of 6π cm2/s
(b) r = 4cm
For the circle, we take radius = r and area =A
Given r= 3 cm, Area of circle A = πr2
Here, we have to find the rate of change of area A w.r.t.’ r ‘.
∴ A = πr2 ⇒ \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = π(2r) = 2πr
When r = 5, \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = 2π(4) = 8π.
Thus, the area of the circle is changing at the rate of 8π cm2/s.
Question 2.
The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Solution:
For the circle, we take radius = r and area = A
We know that A = πr2
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 3cm / s and r= 10 cm
Now \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(πr2) = 2πr \(\frac{\mathrm{dr}}{\mathrm{dt}}\)
= 2π(10)(3) = 60π cm2
Question 3.
An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the Volume of the cube increasing when the edge is 10 cm long?
Solution:
Let x be the length and V be the volume of the cube.
Given that \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 3cm/s ,x = 10cm
Hence, V = x3
Now, \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(x3) = 3x2\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 3(10)2(3) = 900cm3/s
Question 4.
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Solution:
For the circle, we take radius = r and area = A
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 5 and r = 8 Area of circle A = πr2
On diff. w.r.t ‘t’, we get \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = (2πr)\(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 2π(8)(5) = 80π cm2/s
Thus, the enclosed area is increasing at the rate of 80π cm2/s, when r = 8 cm.
Question 5.
The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference? ‘
Solution:
For the circle, we take radius = r and area = A
Given that \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 0.7 cm / s
We know that Circumference C = 2πr
Now, \(\frac{\mathrm{dC}}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(2πr) = 2π\(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 2π(0.7) = 1.4π cm / s
Question 6.
A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
Solution:
For the sphere, we take radius = r and volume = V
Given that radius, r = 10 cm. We know that V = \(\frac{4}{3}\)πr3
∴ \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dr}}\)(\(\frac{4}{3}\)πr2) = \(\frac{4}{3}\)π(3r2) = 4πr2 = 4π(10)2 =400π
Thus, the volume of the balloon is increasing at the rate of 400π cm3/s.
Question 7.
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:
We know that marginal cost is the rate of change of total cost with respect to the output.
∴ cost(MC) = \(\frac{\mathrm{dC}}{\mathrm{dx}}\) = 0.007(3x2) – 0.003(2x) + 15
= 0.021x2 – 0.006x + 15 dx
When x = 17, MC = 0.021(17)2 – 0.006(17) + 15
= 0.021(289) – 0.006(17) + 15
= 6069 – 0102 + 15
= 20.967
Hence, the required marginal cost is ₹ 20.967 (nearly).
Question 8.
The total revenue in Rupees received from the sale of x units of a product is given by R (x) = 13x2 + 26x + 15. Find the marginal revenue when x = 7.
Solution:
Marginal revenue (MR) is die rate of change of the total revenue with respect to the number of units sold.
∴ MR = \(\frac{\mathrm{dR}}{\mathrm{dx}}\) = 13(2x)+26
= 26x + 26
When x = 7, MR = 26(7) + 26 = 182 + 26 = 208
II.
Question 1.
The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?
Solution:
For the cube, we take length of the edge = x , Volume = V and Surface area = S
Given \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 8cm3 / s and x = 12 cm
Volume of the cube V = x3 On diff. w.r.t’t’, we get \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 3x2\(\frac{\mathrm{dx}}{\mathrm{dt}}\)
⇒ 8 = 3x2\(\frac{\mathrm{dx}}{\mathrm{dt}}\) ⇒ \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = \(\frac{8}{3 x^2}\)
Surface area S = 6x2
On diff. w.r.t. ‘t’, we get \(\frac{\mathrm{dS}}{\mathrm{dt}}\) = 12x\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 12x\(\left(\frac{8}{3 x^2}\right)=\frac{32}{x}\)
So, when x = 12 cm ⇒ \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = \(\frac{32}{12}\) = \(\frac{8}{3}\)cm2 / s
Question 2.
The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.
Solution:
Given that \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = -5cm / min , \(\frac{\mathrm{dy}}{\mathrm{dt}}\) = 4cm / min, x = 8cm and y = 6cm
(a) The perimeter of a rectangle is given by P = 2(x+y)
∴ \(\frac{\mathrm{dP}}{\mathrm{dt}}\) = 2\(\left(\frac{\mathrm{dx}}{\mathrm{dt}}+\frac{\mathrm{dy}}{\mathrm{dt}}\right)\) = 2(-5 + 4) = -2 cm / min
(b) The area of a rectangle is given by A = xy
⇒ \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = \(\frac{\mathrm{dx}}{\mathrm{dt}}\)y + x\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = -5y + 4x = (-5 × 6 + 4 × 8)cm2 / min = 2cm2 / min
Question 3.
A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Solution:
For the sphere, we take radius = r and volume = V
Given \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 900c.c/s, r = 15cm
Volume of the sphere V = \(\frac{4}{3}\)πr3
On diff w.r.t ‘t’, we get \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = \(\frac{4}{3}\)π 3 r2 \(\frac{\mathrm{dr}}{\mathrm{dt}}\)
⇒ 900 = 4π(15)2\(\frac{\mathrm{dr}}{\mathrm{dt}}\)
⇒ \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = \(\frac{900}{4 \pi \times 15 \times 15}=\frac{900}{900 \pi}=\frac{1}{\pi}\) cm/s
Question 4.
A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?
Solution:
Let the height of the wall at which the ladder is touching it be y
and the distance of its foot from the wall on the ground be x
Given that x = 4 cm, \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 2cm / s
Hence, x2 + y2 = 52 ⇒ y2 = 25 – x2 ⇒ y = \(\sqrt{25-x^2}\)
∴ \(\frac{d y}{d t}=\frac{d}{d t}\left(\sqrt{25-x^2}\right)=\frac{1}{2 \sqrt{25-x^2}}(-2 x) \frac{d x}{d t}=\frac{-x}{\left(\sqrt{25-x^2}\right)} \frac{d x}{d t}=\frac{-2 x}{\sqrt{25-x^2}}=\frac{-2 \times 4}{\sqrt{25-16}}=-\frac{8}{3}\) cm / s
Question 5.
A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
Solution:
From the question we have \(\frac{\mathrm{dy}}{\mathrm{dt}}\) = 8\(\frac{\mathrm{dx}}{\mathrm{dt}}\)
Given equation of the curve 6y = x3 + 2
Diff. w.r.t time we have 6\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = 3x2\(\frac{\mathrm{dx}}{\mathrm{dt}}\) ⇒ 2\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = x2\(\frac{\mathrm{dy}}{\mathrm{dt}}\)
⇒ 2(8\(\frac{\mathrm{dx}}{\mathrm{dt}}\)) = x2\(\frac{\mathrm{dx}}{\mathrm{dt}}\) ⇒ 16\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = x2\(\frac{\mathrm{dx}}{\mathrm{dt}}\) ⇒ (x2 – 16)\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 0
⇒ x2 = 16 ⇒ x = ±4
When x = 4then y = \(\frac{4^3+2}{6}=\frac{66}{6}\) = 11
When x = -4 then y = \(\frac{\left(-4^3\right)+2}{6}=-\frac{62}{6}=-\frac{31}{3}\)
Thus, the points on the curve are (4, 11) and (-4, \(\frac{-31}{3}\))
Question 6.
The radius of an air bubble is increasing at the rate of \(\frac{1}{2}\) cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?
Solution:
For the sphere, we take radius = r, Volume = V.
Given that \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = \(\frac{1}{2}\)cm / s we have to find \(\frac{\mathrm{dV}}{\mathrm{dt}}\) at r = 1
Volume V = \(\frac{4}{3}\)πr3 ⇒ \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = \(\frac{4}{3}\) π 3 r2\(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 4π(1)2\(\frac{1}{2}\) = 2π cm3/s
Question 7.
A balloon, which always remains spherical, has a variable diameter \(\frac{3}{2}\)(2x + 1). Find the rate of change of its ‘oIurne with respect to x.
Solution:
Given that diameter d = \(\frac{3}{2}\)(2x +1). Hence, radius r = \(\frac{3}{4}\)(2x + 1)
We know that V = \(\frac{4}{3}\)πr3 = \(\frac{4}{3} \pi\left(\frac{3}{4}\right)^3\) (2x + 1)3 = \(\frac{9}{16}\) π(2x + 1)3
∴ \(\frac{\mathrm{dV}}{\mathrm{dx}}\) = \(\frac{9}{16}\) π\(\frac{\mathrm{d}}{\mathrm{dx}}\)(2x + 1)3= \(\frac{9 \pi}{16}\)3(2x + 1)2 . 2
= \(\frac{27}{8}\)π(2x + 1)2
III.
Question 1.
Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
Solution:
For the cone, we take radius = r, Volume = V, height = h
Given that h = \(\frac{1}{6}\)r ⇒ r = 6h, \(\frac{\mathrm{dV}}{\mathrm{dx}}\) = 12cm2 / s and h = 4cm
We know that V = \(\frac{1}{3}\)πr2h.
∴ V = \(\frac{1}{3}\)π(6h)2h = 12πh3
\(\frac{\mathrm{dV}}{\mathrm{dx}}\) = 12π\(\frac{\mathrm{d}}{\mathrm{dh}}\)(h3) = 12π(3h2)\(\frac{\mathrm{dh}}{\mathrm{dt}}\) = 36πh2 \(\frac{\mathrm{dh}}{\mathrm{dt}}\)
h = 4 cm, then 12 = 36π(4)2 \(\frac{\mathrm{dh}}{\mathrm{dt}}\) ⇒ \(\frac{\mathrm{dh}}{\mathrm{dt}}\) = \(\frac{12}{36 \pi(16)}=\frac{1}{48 \pi}\) cm / s