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

«
»

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
advertisement

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.
advertisement
advertisement

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.
advertisement

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
advertisement

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
advertisement

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.

To practice all areas of Mathematics for Interviews, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & technical discussions at Telegram SanfoundryClasses.