# Mathematics Questions and Answers – Derivatives Application – Rate of Change of Quantities

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This set of Mathematics Interview Questions and Answers focuses on “Derivatives Application – Rate of Change of Quantities”.

1. If the rate of change of radius of a circle is 6 cm/s then find the rate of change of area of the circle when r=2 cm.
a) 74.36 cm2/s
b) 75.36 cm2/s
c) 15.36 cm2/s
d) 65.36 cm2/s

Explanation: The rate of change of radius of the circle is $$\frac{dr}{dt}$$=6 cm/s
The area of a circle is A=πr2
Differentiating w.r.t t we get,
$$\frac{dA}{dt}=\frac{d}{dt}$$ (πr2)=2πr $$\frac{dr}{dt}$$=2πr(6)=12πr.
$$\frac{dA}{dt}$$|r=2=24π= 24×3.14=75.36 cm2/s

2. The edge of a cube is increasing at a rate of 7 cm/s. Find the rate of change of area of the cube when x=6 cm.
a) 578 cm2/s
b) 498 cm2/s
c) 504 cm2/s
d) 688 cm2/s

Explanation: Let the edge of the cube be x. The rate of change of edge of the cube is given by $$\frac{dx}{dt}$$=7cm/s.
The area of the cube is A=6x2
∴$$\frac{dA}{dt}=\frac{d}{dt}$$(6x2)=12x.$$\frac{dx}{dt}$$=12x×7=84x
$$\frac{dA}{dt}$$|_x=6=84×6=504 cm2/s.

3. The rate of change of area of a square is 40 cm2/s. What will be the rate of change of side if the side is 10 cm.
a) 2 cm/s
b) 4 cm/s
c) 8 cm/s
d) 6 cm/s

Explanation: Let the side of the square be x.
A=x2, where A is the area of the square
Given that, $$\frac{dA}{dt}$$=2x $$\frac{dx}{dt}$$=40 cm2/s.
$$\frac{dx}{dt}=\frac{20}{x}$$cm/s
$$\frac{dx}{dt}=\frac{20}{10}$$=2 cm/s.

4. The total cost P(x) in rupees associated with a product is given by P(x)=0.4x2+2x-10. Find the marginal cost if the no. of units produced is 5.
a) Rs.3
b) Rs.4
c) Rs.5
d) Rs.6

Explanation: The Marginal cost is the rate of change of revenue w.r.t the no. of units produced, we get
$$\frac{dP(x)}{dt}$$=0.8x+2
cost(MC)=$$\frac{dP(x)}{dt}|$$x=5=0.8x+2=0.8(5)+2=4+2=6.

5. At what rate will the lateral surface area of the cylinder increase if the radius is increasing at the rate of 2 cm/s when the radius is 5 cm and height is 10 cm?
a) 40 cm/s
b) 40π cm/s
c) 400π cm/s
d) 20π cm/s

Explanation: Let r be the radius and h be the height of the cylinder. Then,
$$\frac{dr}{dt}$$=2 cm/s
The area of the cylinder is given by A=2πrh
$$\frac{dA}{dt}$$=2πh($$\frac{dr}{dt}$$)=4πh=4π(10)=40π cm/s.

6. If the circumference of the circle is changing at the rate of 5 cm/s then what will be rate of change of area of the circle if the radius is 6cm.
a) 20 cm2/s
b) 40 cm2/s
c) 70 cm2/s
d) 30 cm2/s

Explanation: The circumference of the circle is given by C=2πr, where r is the radius of the circle.
∴$$\frac{dC}{dt}$$=2π.$$\frac{dr}{dt}$$=5 cm/s
$$\frac{dr}{dt}$$=5/2π cm/s
$$\frac{dA}{dt}$$|r=6=2πr.$$\frac{dr}{dt}$$=2πr.$$\frac{5}{2 \pi}$$=5r=5(6)=30 cm2/s.

7. The total cost N(x) in rupees, associated with the production of x units of an item is given by N(x)=0.06x3-0.01x2+10x-43. Find the marginal cost when 5 units are produced.
a) Rs. 1.44
b) Rs. 144.00
c) Rs. 14.4
d) Rs. 56.2

Explanation: The marginal cost is given by the rate of change of revenue.
Hence, $$\frac{dN(x)}{dt}$$=0.18x2-0.02x+10.
∴$$\frac{dN(x)}{dt}$$|_x=5=0.18(5)2-0.02(5)+10
=4.5-0.1+10
=Rs. 14.4

8. The length of the rectangle is changing at a rate of 4 cm/s and the area is changing at the rate of 8 cm/s. What will be the rate of change of width if the length is 4cm and the width is 1 cm.
a) 5 cm/s
b) 6 cm/s
c) 2 cm/s
d) 1 cm/s

Explanation: Let the length be l, width be b and the area be A.
The Area is given by A=lb
$$\frac{dA}{dt}$$=l.$$\frac{db}{dt}$$+b.$$\frac{dl}{dt}$$ -(1)
Given that, $$\frac{dl}{dt}$$=4cm/s and $$\frac{dA}{dt}$$=8 cm/s
Substituting in the above equation, we get
8=l.$$\frac{db}{dt}$$+4b
Given that, l=4 cm and b=1 cm
∴8=4($$\frac{db}{dt}$$)+4(1)
8=4($$\frac{db}{dt}$$)+4
$$\frac{db}{dt}$$=1 cm/s.

9. For which of the values of x, the rate of increase of the function y=3x2-2x+7 is 4 times the rate of increase of x?
a) -1
b) $$\frac{1}{3}$$
c) 1
d) 0

Explanation: Given that, $$\frac{dy}{dt}=4.\frac{dx}{dt}$$
y=3x2-2x+7
$$\frac{dy}{dt}$$=(6x-2) $$\frac{dx}{dt}$$
4.$$\frac{dx}{dt}$$=(6x-2) $$\frac{dx}{dt}$$
4=6x-2
6x=6
⇒x=1

10. The volume of a cube of edge x is increasing at a rate of 12 cm/s. Find the rate of change of edge of the cube when the edge is 6 cm.
a) $$\frac{1}{8}$$
b) $$\frac{2}{9}$$
c) –$$\frac{1}{9}$$
d) $$\frac{1}{9}$$

Explanation: Let the volume of cube be V.
V=x3
$$\frac{dV}{dt}$$=3x2 $$\frac{dx}{dt}$$
12=3x2 $$\frac{dx}{dt}$$
$$\frac{dx}{dt}=\frac{4}{x^2}$$
$$\frac{dx}{dt}$$|_x=6=$$\frac{4}{6^2}=\frac{4}{36}=\frac{1}{9}$$.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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