The diagonal of a square is changing at the r | vedclass

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

Question

(B) Let $x$ be the length of the diagonal of the square and $A$ be its area.Given that the rate of change of the diagonal is $\frac{dx}{dt} = 0.5 	ex

Vedclass · Accepted Answer

$10 \sqrt{2} 	ext{ cm}^2/	ext{sec}$

Vedclass · Answer

(B) Let $x$ be the length of the diagonal of the square and $A$ be its area.Given that the rate of change of the diagonal is $\frac{dx}{dt} = 0.5 	ext{ cm/sec}$.The area of a square in terms of its diagonal $x$ is given by $A = \frac{x^2}{2}$.Differentiating both sides with respect to time $t$,we get:$\frac{dA}{dt} = \frac{d}{dt} \left( \frac{x^2}{2} ight) = \frac{2x}{2} \cdot \frac{dx}{dt} = x \cdot \frac{dx}{dt}$.Given that the area $A = 400 	ext{ cm}^2$,we find the diagonal $x$ using $A = \frac{x^2}{2}$:$400 = \frac{x^2}{2} \implies x^2 = 800 \implies x = \sqrt{800} = 20\sqrt{2} 	ext{ cm}$.Now,substitute $x = 20\sqrt{2}$ and $\frac{dx}{dt} = 0.5$ into the expression for $\frac{dA}{dt}$:$\frac{dA}{dt} = (20\sqrt{2}) \cdot (0.5) = 10\sqrt{2} 	ext{ cm}^2/	ext{sec}$.

The diagonal of a square is changing at the rate of $0.5 \text{ cm/sec}$. Then the rate of change of area when the area is $400 \text{ cm}^2$ is equal to

Similar Questions

$A$ particle is fired straight up from the ground. Its height in feet after $t$ seconds is given by $s(t) = 128t - 16t^2$. The velocity of the particle when it hits the ground is...

$A$ ladder $5 \ m$ long is leaning against a wall. The bottom of the ladder is pulled along the ground,away from the wall,at the rate of $2 \ cm/s$. How fast is its height on the wall decreasing when the foot of the ladder is $4 \ m$ away from the wall?

The side of a square piece is increasing at a rate of $4 \text{ cm/min}$. Find the rate at which the area is increasing when the side is $8 \text{ cm}$ long.

If the displacement $S$ of a particle travelling along a straight line in $t$ seconds is given by $S = 2t^3 + 2t^2 - 2t - 3$,then the time taken (in seconds) by the particle to change its direction is

Vedclass Test Series

Exam Paper Generator

Online Exam Module