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
View Answer

Answer: b
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
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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
View Answer

Answer: c
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
View Answer

Answer: a
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.
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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
View Answer

Answer: d
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
View Answer

Answer: b
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.
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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
View Answer

Answer: d
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
View Answer

Answer: b
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
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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
View Answer

Answer: d
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
View Answer

Answer: c
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
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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) \(\frac{1}{9}\)
View Answer

Answer: d
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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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter