# Class 12 RD Sharma Solutions- Chapter 13 Derivative as a Rate Measurer – Exercise 13.1

**Question 1. Find the rate of change of the total surface area of a cylinder of radius r and height h, when the radius varies.**

**Solution:**

Let total surface area of the cylinder be A

A = 2πr(r + h)

Now we will differentiating it with respect to r as r varies

dA/dr = 2πr(0+1) + (h+r)2π

dA/dr = 4πr + 2πh

**Question 2. Find the rate of change of the volume of a sphere with respect to its diameter.**

**Solution:**

Let D be the diameter and r be the radius of sphere.

So volume of sphere = 4/3πr^{2}

so we can write as v = 4/24πD^{3}[d = 2r]

Now we will differentiating it with respect to D

dv/dD = 12/24πD^{2}

dv/dD = πD^{2}/2

**Question 3. Find the rate of change of the volume of a sphere with respect to its surface area when the radius is 2 cm.**

**Solution:**

Given in question radius of sphere(r) = 2cm

As we know that, v = 4/3πr^{2}

dv/dr = 4πr^{2}—-(equation i)A = 4πr

^{2}

dA/dr = 8πr^{2}—-(equation ii)

Dividing equation (i) and (ii)

(dv/dr)/(dA/dr) = 4πr^{2}/ 8πr

dv/dA = r/2

dv/dA at r = 2 is 1.

**Question 4. Find the rate of change of the area of a circular disc with respect to its circumference when the radius is 3 cm.**

**Solution:**

Let r be the radius of circular disc.

As we know that Area(A) = πr^{2}

dA/dr = 2πr —(equation i)

circumference(C) = 2πr

dC/dr = 2π —(equation ii)

Dividing equation (i) by (ii)

(dA/dr)/(dc/dr) = 2πr / 2π

dA/dc = r

At r = 3 dA/dc = 3.

**Question 5. Find the rate of change of the volume of a cone with respect to the radius of its base.**

**Solution:**

Let r be the radius

V be the volume of cone

h be the height

As we know that V = 1/3πr^{2}h

dV/dr = 2/3πrh.

**Question 6. Find the rate of change of the area of a circle** **with respect to its radius r when r = 5cm.**

**Solution:**

Let r be the radius

A be the area of circle.

As we know that A = πr^{2}

dA/dr = 2πr

At r=5 , dA/dr = 2π(5)

= 10π

**Question 7. Find the rat of change of the volume of the ball with respect to its radius r. How fast is the volume changing with respect to the radius when the radius is 2cm? **

**Solution:**

Here given in the question , r = 2cm

V = 4/3πr^{3}

dV/dr = 4πr^{2}

At r = 2 , dV/dr = 4π(2)^{2}

= 16π

**Question 8. The total cost C(x) in Rupees associated with the production of x units of an item is given by C(x) = 0.007x**^{3} – 0.003x^{2 }+ 15x + 4000. Find the marginal cost when 17 units are produced.

^{3}– 0.003x

^{2 }+ 15x + 4000. Find the marginal cost when 17 units are produced.

**Solution:**

Here in the given question:

Marginal cost is the rate of change of total cost with respect to output.

Marginal cost(MC) = dC/dx = 0.007(3x^{2}) – 0.003(2x) + 15

= 0.021x^{2}– 0.006x + 15

When x=17 , MC = 0.021(17^{2}) – 0.006(17) + 15

= 6.069 – 0.102 + 15

=20.967

When 17 units are produced , the marginal cost is Rs 20.967.

**Question 9. The total revenue in Rupees received from the sale of x units of a production given by R(x) = 13x**^{2} + 26x + 15. Find the marginal revenue when x = 7.

^{2}+ 26x + 15. Find the marginal revenue when x = 7.

**Solution:**

Marginal revenue is the rate of change of total revenue with respect to the number of units sold

Marginal Revenue(MR) = dR/dx = 13(2x) + 26 = 26x + 26

when x = 7

MR = 26(7) + 26 = 182 +26 = 208

So we can that required marginal cost is Rs208.

**Question 10. The money to be spent for the welfare of the employees of a firm is proportional to the rate of change of its total revenue (Marginal revenue). If the total revenue (in rupees) received from the sale of x units of a product is given by R(x) = 3x**^{2}+36x+5, find the marginal revenue, when x=5,** and write which value does the question indicate.**

^{2}+36x+5, find the marginal revenue, when x=5

**Solution:**

Given function R(x) = 3x

^{2}+ 36x + 5

dR/dx = 6x + 36

At x = 5, dR/dx = 6 x 5 + 36

= 66

According to the question, amount of money spent on welfare of employees.

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