Differential and Integral Calculus Questions and Answers – Rectification

This set of Differential and Integral Calculus Questions and Answers for Freshers focuses on “Rectification”.

1. Rectification is determining ____________
a) Length of a line
b) Length of a curve
c) Area of an object
d) Perimeter of an object
View Answer

Answer: b
Explanation: Rectification is the mathematical process in which the length of a curve is determined by using integral calculus. Here, the curve is segmented into parts of known length and therefore the length is found.

2. Which one of the following is an infinite curve?
a) Hyperbola
b) Koch curve
c) Gaussian curve
d) Parabola
View Answer

Answer: b
Explanation: A curve which has no top limit is the infinite curve. Every arc on the curve has undetermined length. Example: Koch curve.

3. The expression for arc length in rectangular form is_________________
a) \(ds = \int_a^b xy \sqrt{1 + (\frac{dy}{dx})^2} \)
b) \(ds = \int_a^b \sqrt{1 – (\frac{dy}{dx})^2} \,dx \)
c) \(ds = \int_a^b \sqrt{(\frac{dy}{dx})^2} \,dx \)
d) \(ds = \int_a^b \sqrt{1 – (\frac{dy}{dx})^2} \,dx \)
View Answer

Answer: d
Explanation: Rectangular form is represented by x,y and z co-ordinates. The right answer is \(ds = \int_a^b \sqrt{1 – (\frac{dy}{dx})^2} \,dx. \)
advertisement
advertisement

4. The expression for arc length in parametric form is_________________
a) \(ds = \int_a^b xy \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \,ds \)
b) \(ds = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \,dt \)
c) \(ds = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \,dx \)
d) \(ds = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \)
View Answer

Answer: b
Explanation: The parametric equation has more than one dependent variable. The right answer is \(ds = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \,dt \).

5. The expression for arc length in polar form is_________________
a) \(ds = \int_a^b \sqrt{r^2 + (\frac{dr}{dθ})^2} \)
b) \(ds = \int_a^b \sqrt{r^2 + (\frac{dr}{dθ})^2} \,dθ \)
c) \(ds = \int_a^b \sqrt{(\frac{dr}{dθ})^2} \,dθ \)
d) \(ds = \int_a^b \sqrt{r^2 + (\frac{dr}{dθ})^2} \,dr \)
View Answer

Answer: b
Explanation: Polar equation is written in terms of polar coordinates r and θ. The right answer is \(ds = \int_a^b \sqrt{r^2 + (\frac{dr}{dθ})^2} \,dθ \).
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

6. Closed form solutions are absent for ellipses.
a) True
b) False
View Answer

Answer: a
Explanation: Closed form solutions are arrived when only a known number of operations are involved. But in the case of ellipse, it is not obtained.

7. What is the length of a circular curve when θ is in degrees and ‘r’ is the radius?
a) s = rθ
b) s = r
c) \(s = \frac{πrθ}{180} \)
d) s = πr
View Answer

Answer: c
Explanation: Degrees must be converted to equivalent radians and therefore the arc length is found considering both the radius and angle.
advertisement

8. The length of curve r = eθ, θ value ranges between 0 to π is________
a) \(\sqrt{2} \)
b) \(\sqrt{2} \,(e^2π – 1) \)
c) \(\sqrt{2} \,(e^2π + 1) \)
d) \(\sqrt{2} \,(e^2π) \)
View Answer

Answer: b
Explanation: \(s = \int_a^b \sqrt{r^2 + (\frac{dr}{dθ})^2} \,dθ \)
\(= \int_0^{2π} \sqrt{(e^θ)^2 + (e^θ)^2)} \,dθ \)
\(= \int_0^{2π} e^θ \sqrt{2} \,dθ \)
\(= (e^θ \sqrt{2}) |_0^{2π} \)
\(= \sqrt{2} (e^{2π}-1). \)

9. The arc length of y = cosh⁡x where x varies from 0 to 1 is ____________
a) \(\frac{e}{2} – \frac{1}{2e} \)
b) \(\frac{e}{2} – \frac{1}{2e} + \frac{1}{2e} \)
c) \(\frac{e}{2} – \frac{1}{2e} \)
d) 2e
View Answer

Answer: a
Explanation: y’ = sinh⁡x
Arc length = \(\int_a^b \sqrt{1 + (\frac{dy}{dx})^2} \,dx\)
= \(\int_0^1 \sqrt{1 + (sinh⁡x)^2)} \,dx\)
= \(\int_0^1 cosh⁡x \,dx \)
= \(\Big[sinh ⁡x\Big]_0^1\)
= \(\frac{e}{2} – \frac{1}{2e} \).
advertisement

Sanfoundry Global Education & Learning Series – Differential and Integral Calculus.

To practice all areas of Differential and Integral Calculus for Freshers, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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 & discussions at Telegram SanfoundryClasses.