Explore topic-wise InterviewSolutions in Current Affairs.

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1.	In a sphere the rate of change of surface area isA.8π times the rate of change of diameterB.2π times the rate of change of diameterC.2π times the rate of change of radiusD.8π times the rate of change of radius
Answer» Correct answer is D. The surface area of a sphere, with radius r, is defined as A(r) = 4πr2 – (1) Differentiating (1) with respect to t, we get \(\frac{dA}{dt}=8\pi r\frac{dr}{dt}\) = 8π×(current radius)×(rate of change of radius)

In a sphere the rate of change of surface area isA.8π times the rate of change of diameterB.2π times the rate of change of diameterC.2π times the rate of change of radiusD.8π times the rate of change of radius

Answer»

Correct answer is D.

The surface area of a sphere, with radius r, is defined as

A(r) = 4πr2 – (1)

Differentiating (1) with respect to t, we get

\(\frac{dA}{dt}=8\pi r\frac{dr}{dt}\)

= 8π×(current radius)×(rate of change of radius)

Discussion

2.	Find the rate of change of the volume of a sphere with respect to its diameter.
Answer» The volume of a Sphere\(=\frac{1}{6}\pi D^3\) Where D = diameter of the Sphere We need to find,\(\frac{dV}{dD}\)where V = Volume of the sphere and D = Diameter of the Sphere. \(\frac{dV}{dD}=\frac{\pi D^2}{2}\) Hence, Rate of change of Volume of Sphere with respect to the diameter of the Sphere is given by\(\frac{\pi D^2}{2}.\)

Find the rate of change of the volume of a sphere with respect to its diameter.

Answer»

The volume of a Sphere\(=\frac{1}{6}\pi D^3\)

Where D = diameter of the Sphere

We need to find,\(\frac{dV}{dD}\)where V = Volume of the sphere and D = Diameter of the Sphere.

\(\frac{dV}{dD}=\frac{\pi D^2}{2}\)

Hence, Rate of change of Volume of Sphere with respect to the diameter of the Sphere is given by\(\frac{\pi D^2}{2}.\)

Discussion

3.	If x = 2 cos t – cos 2t, y = 2 sin t – sin 2t, find\(\frac{d^2y}{dx^2}\)at t =\(\frac{\pi}{2}\).
Answer» Formula: – Given: – x = 2cost – cos2t y = 2sint – sin2t differentiating w.r.t t

If x = 2 cos t – cos 2t, y = 2 sin t – sin 2t, find\(\frac{d^2y}{dx^2}\)at t =\(\frac{\pi}{2}\).

Answer»

Formula: –

Given: –

x = 2cost – cos2t

y = 2sint – sin2t

differentiating w.r.t t

Discussion

4.	Find \(\frac{d^2y}{dx^2}\), where y = log \((\frac{x^2}{e^2})\).
Answer» Formula: – Again Differentiating w.r.t x

Find \(\frac{d^2y}{dx^2}\), where y = log \((\frac{x^2}{e^2})\).

Answer»

Formula: –

Again Differentiating w.r.t x

Discussion

5.	If y = 3 e2x + 2 e3x, prove that \(\frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y\) = 0.
Answer» Formula: – Given: – y = 3e2x + 2e3x Hence

If y = 3 e2x + 2 e3x, prove that \(\frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y\) = 0.

Answer»

Formula: –

Given: –

y = 3e2x + 2e3x

Hence

Discussion

6.	If x = cos t + log tan \(\frac{t}{2}\), y = sin t, then find the value of \(\frac{d^2y}{dt^2}and\frac{d^2y}{dx^2}\).at t =\(\frac{\pi}{4}\).
Answer» Formula: – Given: – x = cos t + log tan \(\frac{t}{2}\), y = sin t Differentiating with respect to t ,we have We have

If x = cos t + log tan \(\frac{t}{2}\), y = sin t, then find the value of \(\frac{d^2y}{dt^2}and\frac{d^2y}{dx^2}\).at t =\(\frac{\pi}{4}\).

Answer»

Formula: –

Given: –

x = cos t + log tan \(\frac{t}{2}\), y = sin t

Differentiating with respect to t ,we have

We have

Discussion

7.	If x = 3 cot t – 2 cos3 t, y = 3 sin t – 2 sin3 t, find\(\frac{d^2y}{dx^2}\)
Answer» Formula: – given: – x = 3 cot t – 2 cos3 t, y = 3 sin t – 2 sin3 t differentiating both w.r.t t

If x = 3 cot t – 2 cos3 t, y = 3 sin t – 2 sin3 t, find\(\frac{d^2y}{dx^2}\)

Answer»

Formula: –

given: –

x = 3 cot t – 2 cos3 t, y = 3 sin t – 2 sin3 t

differentiating both w.r.t t

Discussion

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