The rate of change of the diagonal R of a squ | vedclass

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

Question

(B) Let the side length of the square be $x$.The diagonal $R$ is given by $R = \sqrt{x^2 + x^2} = x\sqrt{2}$.The area $A$ is given by $A = x^2$.From t

Vedclass · Accepted Answer

$1/R$

Vedclass · Answer

(B) Let the side length of the square be $x$.The diagonal $R$ is given by $R = \sqrt{x^2 + x^2} = x\sqrt{2}$.The area $A$ is given by $A = x^2$.From the area equation,we have $x = \sqrt{A}$.Substituting this into the diagonal equation,we get $R = \sqrt{2} \cdot \sqrt{A} = \sqrt{2A}$.To find the rate of change of $R$ with respect to $A$,we differentiate $R$ with respect to $A$:$\frac{dR}{dA} = \frac{d}{dA}(\sqrt{2} \cdot A^{1/2}) = \sqrt{2} \cdot \frac{1}{2} A^{-1/2} = \frac{\sqrt{2}}{2\sqrt{A}} = \frac{1}{\sqrt{2A}}$.Since $R = \sqrt{2A}$,we can substitute this back into the expression:$\frac{dR}{dA} = \frac{1}{R}$.

The rate of change of the diagonal $R$ of a square with respect to its area $A$ is:

Similar Questions

$A$ particle is in motion along a curve $12 y = x^{3}$. The rate of change of its ordinate exceeds that of its abscissa when:

$A$ stone is dropped into a quiet lake and waves move in circles at a speed of $8 \,cm/sec$. At the instant when the radius of the circular wave is $12 \,cm$, how fast is the enclosed area increasing?

Gas is being pumped into a spherical balloon at the rate of $30 \ ft^3 / \text{min}$. The rate at which the radius increases when it reaches the value $15 \ ft$,is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module