A gardener is digging a plot of land. As he g | vedclass

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

Question

(C) Given the rate of digging $\frac{dA}{dt} = \frac{2}{\sqrt{t}}$.Integrating both sides with respect to $t$:$A = \int \frac{2}{\sqrt{t}} dt = 2 \int

Vedclass · Accepted Answer

$100$ minutes

Vedclass · Answer

(C) Given the rate of digging $\frac{dA}{dt} = \frac{2}{\sqrt{t}}$.Integrating both sides with respect to $t$:$A = \int \frac{2}{\sqrt{t}} dt = 2 \int t^{-1/2} dt = 2 \cdot \frac{t^{1/2}}{1/2} + C = 4\sqrt{t} + C$.At $t = 0$,the area dug $A = 0$,so $C = 0$.Thus,$A = 4\sqrt{t}$.To find the time $t$ to dig $40$ square metres,set $A = 40$:$40 = 4\sqrt{t} \Rightarrow \sqrt{t} = 10$.Squaring both sides,$t = 10^2 = 100$ minutes.

$A$ gardener is digging a plot of land. As he gets tired,he works more slowly. After $t$ minutes,he is digging at a rate of $\frac{2}{\sqrt{t}}$ square metres per minute. How long will it take him to dig an area of $40$ square metres?

Similar Questions

Find an anti-derivative (or integral) of the function $e^{2x}$ by the method of inspection.

Find the integral of the function $\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}$.

$\int \frac{e^{2030 \log x}-e^{2029 \log x}}{e^{2028 \log x}-e^{2027 \log x}} \,d x = \dots$

Find the integral of the function $\cos 2x \cos 4x \cos 6x$.

$\int \frac{dx}{4x^2 + 9} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module