Mathematics Interview Questions and Answers 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
Answer: b
Clarification: 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
Answer: c
Clarification: 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
Answer: a
Clarification: 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
Answer: d
Clarification: 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
Answer: b
Clarification: 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
Answer: d
Clarification: 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
Answer: b
Clarification: 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
Answer: d
Clarification: 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
Answer: c
Clarification: 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})
Answer: d
Clarification: 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}).
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