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

The relation between pressure $p$ and volume $V$ is given by $p V^{1/4} = \text{constant}$. If the percentage decrease in volume is $\frac{1}{2} \%$, then the percentage increase in pressure is:

$A$ cylindrical tank of radius $10 \ m$ is being filled with wheat at the rate of $314 \ m^3/h$. The rate at which the depth of the wheat is increasing is: (in $m/h$)

The displacement $S$ of a moving particle at a time $t$ is given by $S=5+\frac{48}{t}+t^3$. Then its acceleration when the velocity is zero,is

$A$ spherical balloon is being inflated at the rate of $35 \, cc/min$. The rate of increase of the surface area of the balloon when its diameter is $14 \, cm$ is ....... $sq \, cm/min$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module