The side of a square is increasing at the rat | vedclass

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

Question

(D) Let the side of the square be $a$. The perimeter $P$ of a square is given by $P = 4a$.To find the rate of increase of the perimeter with respect t

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(D) Let the side of the square be $a$. The perimeter $P$ of a square is given by $P = 4a$.To find the rate of increase of the perimeter with respect to time $t$,we differentiate $P$ with respect to $t$:$\frac{dP}{dt} = \frac{d}{dt}(4a) = 4 \left( \frac{da}{dt} ight)$.Given that the rate of increase of the side is $\frac{da}{dt} = 0.2\,cm/s$.Substituting this value into the equation:$\frac{dP}{dt} = 4 	imes 0.2 = 0.8\,cm/s$.Therefore,the rate of increase of the perimeter is $0.8\,cm/s$.

The side of a square is increasing at the rate of $0.2\,cm/s$. The rate of increase of perimeter with respect to time is $...........\,cm/s$.

Similar Questions

If $xy = c^2$ where $c$ is a constant,then $\frac{dy}{dx}$ is equal to

What is the behavior of the magnitude of the slope of the shown graph?

The sum of the series $1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots \infty$ is

${d \over {dx}}({x^2}{e^x}\sin x) = $

Which of the following relation$(s)$ is/are correct?
$(A)\ \sin (90^{\circ}+\theta)=\cos (-\theta)$
$(B)\ \sin (180^{\circ}-\theta)=\cos (90^{\circ}-\theta)$
$(C)\ \sin (360^{\circ}-\theta)=\cos (360^{\circ}-\theta)$
$(D)\ \sin (180^{\circ}+\theta)=\cos (90^{\circ}-\theta)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module