10.3 Rate Measure (SAQs)
Maths-1B | 10.3 Rate Measure – SAQs:
Welcome to SAQs in Chapter 10: 10.3 Rate Measure. This page contains the most Important FAQs for Short Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.
SAQ-1 : A particle is moving in a straight line so that after t seconds its distances (in cms) from a fixed point on the line is given by s = f(t) = 8t + t3. Find (i) the velocity at time t = 2 sec (ii) the initial velocity (iii) acceleration at t = 2sec
$$s = f(t) = 8t + t^3 \quad \text{(1)}$$
$$v = \frac{ds}{dt} = 8 + 3t^2 \quad \text{(2)}$$
$$a = \frac{dv}{dt} = 6t \quad \text{(3)}$$
(i) Velocity at t=2:
$$v = 8 + 3(2^2) = 8 + 3(4) = 20 \, \text{cm/sec}$$
(ii) Initial velocity at t=0:
$$v = 8 + 3(0) = 8 \, \text{cm/sec}$$
(iii) Acceleration at t=2:
$$a = 6(2) = 12 \, \text{cm/sec}^2$$
SAQ-2 : A particle is moving along a line according s = f(t) = 4t3 – 3t2 + 5t – 1 where s is measured in meters and t is measured in seconds. Find the velocity and acceleration at time t. AT what time the acceleration is zero
Given Position Function and Derivatives:
$$s = f(t) = 4t^3 – 3t^2 + 5t – 1 \quad \text{(1)}$$
$$v = \frac{ds}{dt} = 12t^2 – 6t + 5 \quad \text{(2)}$$
$$a = \frac{dv}{dt} = 24t – 6 \quad \text{(3)}$$
$$24t – 6 = 0$$
$$24t = 6 \
t = \frac{6}{24} = \frac{1}{4}$$
SAQ-3 : A stone is dropped into a quiet lake and ripples move in circles at the speed of 5 cm/sec. At the instant when the radius of circular ripple is 8cm, how fast is the enclosed area increases?
$$\frac{dr}{dt} = 5 \quad \text{cm/sec} \quad \text{and} \quad r = 8 \quad \text{cm}$$
$$A = \pi r^2$$
$$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$$
$$\frac{dA}{dt} = 2\pi (8)(5) = 80\pi \quad \text{cm}^2/\text{sec}$$
SAQ-4 : The radius of an air bubble is increasing at the rate of 1/2 cm/sec. At what rate is the volume of the bubble increasing when the radius is 1cm?
Given Information About Sphere’s Radius and Volume:
$$\frac{dr}{dt} = \frac{1}{2} \quad \text{cm/sec}$$
$$V = \frac{4}{3} \pi r^3$$
$$\frac{dV}{dt} = \frac{d}{dt} \left(\frac{4}{3} \pi r^3\right)$$
$$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$
$$\frac{dV}{dt} = 4\pi (1)^2 \left(\frac{1}{2}\right) = 2\pi \quad \text{cm}^3/\text{sec}$$
SAQ-5 : The volume of a cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of the edge is 10 centimeters?
Given Information About Cube’s Edge Length, Volume, and Surface Area:
$$\frac{dV}{dt} = 9 \quad \text{cm}^3/\text{sec}$$
$$x = 10 \quad \text{cm}$$
$$V = x^3$$
$$\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$$
$$9 = 3(10)^2 \frac{dx}{dt}$$
$$\frac{dx}{dt} = \frac{9}{3 \times 10^2} = \frac{3}{100} = 0.03 \quad \text{cm/sec}$$
$$S = 6x^2$$
$$\frac{dS}{dt} = 12x \frac{dx}{dt}$$
$$\frac{dS}{dt} = 12 \times 10 \times 0.03 = 3.6 \quad \text{cm}^2/\text{sec}$$
SAQ-6 : The volume of a cube is increasing at a rate of 8 cubic centimeters per second. How fast is the surface area increasing when length of edge is 12cm?
Given Cube Parameters and Their Rates of Change:
$$\frac{dV}{dt} = 8 \quad \text{cm}^3/\text{sec}, \quad x = 12 \quad \text{cm}$$
$$V = x^3$$
$$\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$$
$$8 = 3 \times 12^2 \times \frac{dx}{dt}$$
$$\frac{dx}{dt} = \frac{8}{3 \times 12^2} = \frac{8}{432} = \frac{1}{54} \quad \text{cm/sec}$$
$$S = 6x^2$$
$$\frac{dS}{dt} = 12x \frac{dx}{dt}$$
$$\frac{dS}{dt} = 12 \times 12 \times \frac{1}{54} = \frac{144}{54} = \frac{8}{3} \quad \text{cm}^2/\text{sec}$$
SAQ-7 : A balloon which always remains spherical on inflation is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of balloon increases when the radius in 15 cm
We take r as radius and V as volume of the sphere
$$\frac{dV}{dt} = 900 \quad \text{c.c/sec}, \quad r = 15 \quad \text{cm}$$
$$V = \frac{4}{3} \pi r^3$$
$$\frac{dV}{dt} = \frac{d}{dt} \left(\frac{4}{3} \pi r^3\right) = 4 \pi r^2 \frac{dr}{dt}$$
$$900 = 4 \pi (15)^2 \frac{dr}{dt}$$
$$\frac{dr}{dt} = \frac{900}{4 \pi \times 15 \times 15} = \frac{900}{900 \pi} = \frac{1}{\pi} \quad \text{cm/sec}$$