Here we provide RD Sharma Class 12 Ex 13.1 Solutions Chapter 13 Derivative as a Rate Measurer for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 13.1 Solutions Chapter 13 Derivative as a Rate Measurer book pdf download. Now you will get step-by-step solutions to each question.
RD Sharma Class 12 Ex 13.1 Solutions Chapter 13 Derivative as a Rate Measurer
Question 1. The side of a square sheet is increasing at the rate of 4 cm per
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.
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.
Let D be the diameter and r be the radius of sphere.
So volume of sphere = 4/3πr2
so we can write as v = 4/24πD3 [d = 2r]
Now we will differentiating it with respect to D
dv/dD = 12/24πD2
dv/dD = πD2/2
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.
Given in question radius of sphere(r) = 2cm
As we know that, v = 4/3πr2
dv/dr = 4πr2 —-(equation i)
A = 4πr2
dA/dr = 8πr2 —-(equation ii)
Dividing equation (i) and (ii)
(dv/dr)/(dA/dr) = 4πr2 / 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.
Let r be the radius of circular disc.
As we know that Area(A) = πr2
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.
Let r be the radius
V be the volume of cone
h be the height
As we know that V = 1/3πr2h
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.
Let r be the radius
A be the area of circle.
As we know that A = πr2
dA/dr = 2πr
At r=5 , dA/dr = 2π(5)
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?
Here given in the question , r = 2cm
V = 4/3πr3
dV/dr = 4πr2
At r = 2 , dV/dr = 4π(2)2
Question 8. The total cost C(x) in Rupees associated with the production of x units of an item is given by C(x) = 0.007x3 – 0.003x2 + 15x + 4000. Find the marginal cost when 17 units are produced.
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(3x2) – 0.003(2x) + 15
= 0.021x2 – 0.006x + 15
When x=17 , MC = 0.021(172) – 0.006(17) + 15
= 6.069 – 0.102 + 15
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) = 13x2 + 26x + 15. Find the marginal revenue when x = 7.
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) = 3x2+36x+5, find the marginal revenue, when x=5, and write which value does the question indicate.
Given function R(x) = 3x2 + 36x + 5
dR/dx = 6x + 36
At x = 5, dR/dx = 6 x 5 + 36
According to the question, amount of money spent on welfare of employees.
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