x and y are the sides of two squares such tha | vedclass

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

Question

(A) Let $A_{1}$ be the area of the first square and $A_{2}$ be the area of the second square.Given that the side of the first square is $x$,so $A_{1}

Vedclass · Accepted Answer

$2x^{2} - 3x + 1$

Vedclass · Answer

(A) Let $A_{1}$ be the area of the first square and $A_{2}$ be the area of the second square.Given that the side of the first square is $x$,so $A_{1} = x^{2}$.The side of the second square is $y = x - x^{2}$,so $A_{2} = y^{2} = (x - x^{2})^{2}$.We need to find the rate of change of $A_{2}$ with respect to $A_{1}$,which is $\frac{dA_{2}}{dA_{1}}$.Using the chain rule,$\frac{dA_{2}}{dA_{1}} = \frac{dA_{2}/dx}{dA_{1}/dx}$.First,calculate $\frac{dA_{1}}{dx} = \frac{d}{dx}(x^{2}) = 2x$.Next,calculate $\frac{dA_{2}}{dx} = \frac{d}{dx}(x - x^{2})^{2} = 2(x - x^{2}) \cdot \frac{d}{dx}(x - x^{2}) = 2(x - x^{2})(1 - 2x)$.Now,$\frac{dA_{2}}{dA_{1}} = \frac{2(x - x^{2})(1 - 2x)}{2x} = \frac{2x(1 - x)(1 - 2x)}{2x}$.Simplifying this,we get $\frac{dA_{2}}{dA_{1}} = (1 - x)(1 - 2x) = 1 - 2x - x + 2x^{2} = 2x^{2} - 3x + 1$.

$x$ and $y$ are the sides of two squares such that $y = x - x^{2}$. Find the rate of change of the area of the second square with respect to the area of the first square.

Similar Questions

Let $x$ and $y$ be the sides of two squares such that $y = x - x^2$. The rate of change of the area of the second square with respect to the area of the first square is

$A$ particle moves such that $x = 2 + 27t - t^3$. The direction of motion reverses after moving a distance of ... units.

$A$ water tank has the shape of an inverted right circular cone,whose semi-vertical angle is $\tan^{-1}(1/2)$. Water is poured in at a constant rate of $5 \ m^3/min$. Then the rate (in $m/min$) at which the level of water is rising at the instant when the depth of water in the tank is $10 \ m$ is:

From a balloon rising vertically with a uniform velocity of $v \ ft/sec$,a stone is dropped. If the stone reaches the ground after $4 \ sec$,what is the height of the balloon above the ground at that moment (in $ft$)? (Take $g = 32 \ ft/sec^2$)

$A$ man,$2 \ m$ tall,walks at the rate of $1 \frac{2}{3} \ m/s$ towards a street light which is $5 \frac{1}{3} \ m$ above the ground. At what rate is the tip of his shadow moving? At what rate is the length of the shadow changing when he is $3 \frac{1}{3} \ m$ from the base of the light?

Vedclass Test Series

Exam Paper Generator

Online Exam Module