(A) The area $A$ of a circular plate with radius $r$ is given by $A = \pi r^2$. To find the rate at which the area increases,we differentiate $A$ with

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

$0.24 \pi \text{ cm}^2/\text{s}$

(A) The area $A$ of a circular plate with radius $r$ is given by $A = \pi r^2$. To find the rate at which the area increases,we differentiate $A$ with respect to time $t$: $\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2 \pi r \frac{dr}{dt}$. Given that $\frac{dr}{dt} = 0.01 \text{ cm/s}$ and $r = 12 \text{ cm}$,we substitute these values into the equation: $\frac{dA}{dt} = 2 \pi (12) (0.01) = 0.24 \pi \text{ cm}^2/\text{s}$. Thus,the rate at which the area increases is $0.24 \pi \text{ cm}^2/\text{s}$.

Class 12 Mathematics Application of Derivatives Rate of Change of Quantities

Difficult

The radius of a circular plate is increasing at the rate of $0.01 \text{ cm/s}$ when the radius is $12 \text{ cm}$. Then,the rate at which the area increases,is

AP EAMCET 2005 All AP EAMCET

TS EAMCET 2005 All TS EAMCET

A
$0.24 \pi \text{ cm}^2/\text{s}$
B
$60 \pi \text{ cm}^2/\text{s}$
C
$24 \pi \text{ cm}^2/\text{s}$
D
$1.2 \pi \text{ cm}^2/\text{s}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

Application of Derivatives

Subtopic

Rate of Change of Quantities

Previous Year

AP EAMCET 2005 Paper All AP EAMCET years →

Previous Year

TS EAMCET 2005 Paper All TS EAMCET years →

If the rate of change of the slope of the tangent drawn to the curve $y=x^3-2x^2+3x-2$ at the point $(2,4)$ is $k$ times the rate of change of its abscissa, then $k=$

MediumTS EAMCET 2025

View Solution

$A$ spherical balloon is being inflated at the rate of $35 \, cm^3/min$. The rate of increase in the surface area (in $cm^2/min$) of the balloon when its diameter is $14 \, cm$,is

DifficultJEE MAIN 2013

View Solution

The radius of a circle is increasing at the rate of $2 \text{ cm/sec}$. The rate at which its area is increasing when the radius of the circle is $5 \text{ decimeters}$ is:

EasyMHT CET 2020

View Solution

$A$ particle moves in a straight line so that its velocity at any point is given by ${v^2} = a + bx$,where $a, b \neq 0$ are constants. The acceleration is

Easy

View Solution

$A$ point is moving along the curve $y^3 = 27x$. The interval in which the abscissa changes at a slower rate than the ordinate is

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The radius of a circular plate is increasing at the rate of $0.01 \text{ cm/s}$ when the radius is $12 \text{ cm}$. Then,the rate at which the area increases,is

Similar Questions

If the rate of change of the slope of the tangent drawn to the curve $y=x^3-2x^2+3x-2$ at the point $(2,4)$ is $k$ times the rate of change of its abscissa, then $k=$

$A$ spherical balloon is being inflated at the rate of $35 \, cm^3/min$. The rate of increase in the surface area (in $cm^2/min$) of the balloon when its diameter is $14 \, cm$,is

The radius of a circle is increasing at the rate of $2 \text{ cm/sec}$. The rate at which its area is increasing when the radius of the circle is $5 \text{ decimeters}$ is:

$A$ particle moves in a straight line so that its velocity at any point is given by ${v^2} = a + bx$,where $a, b \neq 0$ are constants. The acceleration is

$A$ point is moving along the curve $y^3 = 27x$. The interval in which the abscissa changes at a slower rate than the ordinate is

Vedclass Test Series

Exam Paper Generator

Online Exam Module