Q.$x$ 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.

Showing the 12 most recent of 14 on this concept.

1. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.If $y=x-x^2$, then the rate of change of $y^2$ with respect to $x^2$ at $x=2$ is (A) $0$ (B) $-1$ (C) $3$ (D) $9$

   ›Reveal solutionSolution

   "Rate of change of $y^2$ w.r.t. $x^2$" means $\dfrac{d(y^2)}{d(x^2)}$, computed by dividing derivatives with respect to $x$.

   Concept and Intuition

   When asked for the derivative of one quantity with respect to another (neither being the independent variable $x$), use $\dfrac{du}{dv}=\dfrac{du/dx}{dv/dx}$. Here $u=y^2$, $v=x^2$, both expressed through the parameter $x$.

   Step-by-Step Solution

   1. $y=x-x^2 \Rightarrow y'=1-2x$.
   2. $\dfrac{d(y^2)}{dx}=2yy'$ and $\dfrac{d(x^2)}{dx}=2x$.
   3. So $\dfrac{d(y^2)}{d(x^2)}=\dfrac{2yy'}{2x}=\dfrac{yy'}{x}$.
   4. At $x=2$: $y=2-4=-2$, and $y'=1-4=-3$. …

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2. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.The rate of change of $x^{\sin x}$ with respect to $(\sin x)^x$ is (A) $\dfrac{x^{\sin x}\left(\dfrac{\sin x}{x} + \cos x \cdot \log x\right)}{(\sin x)^x (x \cdot \cot x + \log \sin x)}$ (B) $\dfrac{(\sin x)^x (x \cot x + \log \sin x)}{x^{\sin x}\left(\dfrac{\sin x}{x} + \cos x \cdot \log x\right)}$ (C) $y\left(\dfrac{\sin x}{x} + \cos x \cdot \log x\right)$ (D) $(\sin x)^x (x \cot x + \log \sin x)$

   ›Reveal solutionSolution

   The "rate of change of $y$ w.r.t. $z$" means $\dfrac{dy}{dz}=\dfrac{dy/dx}{dz/dx}$; use logarithmic differentiation on each variable-exponent function.

   Concept and Intuition

   For functions of the form $(\text{variable})^{\text{variable}}$, logarithmic differentiation converts the product/power mess into a clean sum, since $\log y = (\text{exponent})\log(\text{base})$ turns multiplication into differentiable products.

   Step-by-Step Solution

   1. Let $y=x^{\sin x}$. Take logs: $\log y=\sin x\log x$.
   2. Differentiate: $\dfrac{y'}{y}=\cos x\log x+\dfrac{\sin x}{x}$, so $y'=x^{\sin x}\left(\dfrac{\sin x}{x}+\cos x\log x\right)$.
   3. Let $z=(\sin x)^x$. Take logs: $\log z=x\log(\sin x)$.
   4. Differentiate: $\dfrac{z'}{z}=\log(\sin x)+x\cdot\dfrac{\cos x}{\sin x}=\log\sin x+x\cot x$, so $z'=(\sin x)^x(x\cot x+\log\sin x)$. …

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3. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.For the curve $y=5x-2x^3$, if '$x$' increases at the rate of 2 units/sec, the rate of change in the slope of the curve at $x=3$ is ________ /sec (A) 72 (B) 27 (C) -72 (D) -27

   ›Reveal solutionSolution

   Differentiating the slope expression with respect to time (via the chain rule) and plugging in $x=3$, $dx/dt=2$ gives $-72$ per second.

   Concept and Intuition

   This is a related-rates problem: the "slope of the curve" is itself a function of $x$, namely $dy/dx$. Since $x$ changes with time, the slope also changes with time, and we find that rate via the chain rule: $\dfrac{d(\text{slope})}{dt} = \dfrac{d(\text{slope})}{dx}\cdot\dfrac{dx}{dt}$.

   Step-by-Step Solution

   1. $y = 5x-2x^3$, so slope $m = \dfrac{dy}{dx} = 5-6x^2$.
   2. Differentiate $m$ with respect to $x$: $\dfrac{dm}{dx} = -12x$.
   3. By the chain rule, $\dfrac{dm}{dt} = \dfrac{dm}{dx}\cdot\dfrac{dx}{dt} = -12x\cdot\dfrac{dx}{dt}$. …

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4. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.Given $f(x)=x^3-4x$, if '$x$' changes from 2 to 1.99, then the approximate change in the value of $f(x)$ is ______. (A) $0.08$ (B) $-0.08$ (C) $0.8$ (D) $-0.8$

   ›Reveal solutionSolution

   Use the linear (differential) approximation $\Delta f\approx f'(x)\Delta x$. Answer: $-0.08$.

   Concept and Intuition

   For small changes, $\Delta f\approx f'(x)\,\Delta x$ — this is the first-order Taylor/differential approximation, exactly what "approximate change" is asking for.

   Step-by-Step Solution

   1. $f(x)=x^3-4x\Rightarrow f'(x)=3x^2-4$.
   2. At $x=2$: $f'(2)=3(4)-4=8$.
   3. $\Delta x=1.99-2=-0.01$. …

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5. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.The radius and the height of a right circular solid cone are measured as 7 feet each. If there is an error of 0.002 ft for every feet in measuring them, then the error in the total surface area of the cone (in sq. ft) is (A) $(0.088)(\sqrt{2}+1)$ (B) $(0.616)(\sqrt{2}+1)$ (C) $(0.616)(\sqrt{2})$ (D) $(0.088)(\sqrt{2})$

   ›Reveal solutionSolution

   Propagate the proportional measurement error ($0.002$ per foot on both $r$ and $h=7$) through the total-surface-area formula of a cone; the error works out to $0.196\pi(\sqrt2+1)\approx0.616(\sqrt2+1)$ sq ft.

   Concept and Intuition

   Errors in measured quantities propagate to derived quantities (like surface area) via the total differential: if $S=S(r,h)$, then $dS \approx \dfrac{\partial S}{\partial r}\Delta r + \dfrac{\partial S}{\partial h}\Delta h$. Here the slant height $l=\sqrt{r^2+h^2}$ also depends on both $r$ and $h$, so we must differentiate it too.

   Step-by-Step Solution

   1. Total surface area of a right circular cone: $S=\pi r l + \pi r^2$, where $l=\sqrt{r^2+h^2}$ is the slant height.
   2. Given $r=h=7$ ft, so $l=\sqrt{49+49}=7\sqrt2$ ft.
   3. Error rate: $0.002$ ft per foot measured, so $\Delta r = \Delta h = 0.002\times7=0.014$ ft.
   4. Differentiate $S$: $dS=\pi\,l\,dr+\pi r\,dl+2\pi r\,dr$, where $dl=\dfrac{r\,dr+h\,dh}{l}$ (from differentiating $l^2=r^2+h^2$).
   5. Since $r=h$ and $\Delta r=\Delta h$, by symmetry $dl=\dfrac{r\Delta r + r\Delta r}{r\sqrt2}=\dfrac{2r\Delta r}{r\sqrt2}=\Delta r\sqrt2$.
   6. Substitute back: $dS = \pi(r\sqrt2)\Delta r + \pi r(\Delta r\sqrt2)+2\pi r\Delta r = 2\pi r\Delta r\sqrt2 + 2\pi r\Delta r = 2\pi r\Delta r(\sqrt2+1).$ …

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6. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.If the radius of a sphere is measured as $9\ cm$ with an error of $0.03\ cm$, then find the approximate error in calculating its surface area. (A) $2.16\pi\ cm^2$ (B) $21.6\pi\ cm^2$ (C) $216\pi\ cm^2$ (D) $0.216\pi\ cm^2$

   ›Reveal solutionSolution

   This tests using differentials to approximate the propagated error in surface area from a small error in radius. Answer: $2.16\pi\ \text{cm}^2$.

   Concept and Intuition

   For small errors, $\Delta S \approx dS = \dfrac{dS}{dr}\,dr$, turning a nonlinear error-propagation problem into simple differentiation and substitution.

   Step-by-Step Solution

   1. Surface area of a sphere: $S = 4\pi r^2$.
   2. $\dfrac{dS}{dr} = 8\pi r$.
   3. Given $r=9$ cm and $dr = 0.03$ cm: $dS = 8\pi(9)(0.03)$.
   4. $8\times 9 = 72$; $72\times 0.03 = 2.16$. …

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7. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The surface area of a cube is 150 sq. cm. If it is increased by 0.025 sq. cm, then the approximate increase in its volume (in c.c.) is (A) 0.0725 (B) 0.04 (C) 0.032 (D) 0.03125

   ›Reveal solutionSolution

   This tests using differentials to propagate a small change in surface area to a small change in volume via the common edge length; the answer is $0.03125$ c.c.

   Concept and Intuition

   Both the surface area $S=6x^2$ and volume $V=x^3$ of a cube depend only on the edge $x$. A small change $dS$ produces a corresponding small change $dx$ (via $dS=12x\,dx$), and that same $dx$ then produces $dV=3x^2dx$ — this is the standard differentials/approximation technique.

   Step-by-Step Solution

   1. $S=6x^2=150 \Rightarrow x^2=25 \Rightarrow x=5$ cm.
   2. $dS = 12x\,dx \Rightarrow 0.025 = 12(5)\,dx = 60\,dx \Rightarrow dx = \dfrac{0.025}{60} = \dfrac{1}{2400}$ cm. …

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8. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If the percentage error in the radius of circle is 3, then the percentage error in its area is (A) $6$ (B) $\dfrac{3}{2}$ (C) $2$ (D) $4$

   ›Reveal solutionSolution

   For $A=\pi r^2$, percentage error in area = $2\times$ percentage error in radius $= 2\times 3=6\%$.

   Concept and Intuition

   When a quantity is a power of another, $A=kr^n$, small relative errors combine as $\dfrac{dA}{A}=n\dfrac{dr}{r}$. Here $n=2$.

   Step-by-Step Solution

   1. $A=\pi r^2 \Rightarrow dA = 2\pi r\, dr$.
   2. $\dfrac{dA}{A} = \dfrac{2\pi r\,dr}{\pi r^2} = 2\dfrac{dr}{r}$. …

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9. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.If the error committed in measuring the radius of a circle is 0.05 %, then the corresponding error in calculating its area would be ____ (A) 0.05 % (B) 0.0025 % (C) 0.25 % (D) 0.1 %

   ›Reveal solutionSolution

   This tests error propagation through a power-law formula: for $A \propto r^n$, the percentage error in $A$ is $n$ times the percentage error in $r$. Here $n=2$, so the area error is double the radius error.

   Concept and Intuition

   When a measured quantity is used in a formula, small errors propagate. If $A = k r^n$ for a constant $k$ and integer power $n$, taking logarithms gives $\log A = \log k + n\log r$. Differentiating, $\dfrac{dA}{A} = n\dfrac{dr}{r}$. This is the standard rule: percentage error in a power gets multiplied by that power.

   Step-by-Step Solution

   1. Area of a circle: $A = \pi r^2$.
   2. Taking logarithmic differential: $\dfrac{dA}{A} = 2\dfrac{dr}{r}$. …

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10. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.The semi-vertical angle of a right circular cone is $45°$. If the radius of the base of the cone is measured as 14 cm with an error of $\left(\dfrac{\sqrt{2}-1}{11}\right)$cm, then the approximate error in measuring its total surface area is (in sq. cm) (A) $14$ (B) $8$ (C) $5$ (D) $3$

    ›Reveal solutionSolution

    Use differentials to propagate the radius error through the cone's total surface area formula; $dS\approx8$ sq. cm.

    Concept and Intuition

    Small errors propagate through a formula $S=S(r)$ via $dS\approx S'(r)\,dr$ — the differential approximation. Here we first need $S$ purely in terms of $r$, using the given semi-vertical angle to eliminate the slant height and height.

    Step-by-Step Solution

    1. Semi-vertical angle $\theta=45°$ means $\tan\theta=\dfrac{r}{h}=1\Rightarrow h=r$, and $\sin\theta=\dfrac{r}{l}\Rightarrow l=\dfrac{r}{\sin45°}=r\sqrt2$.
    2. Total surface area: $S=\pi r^2+\pi r l=\pi r^2+\pi r(r\sqrt2)=\pi r^2(1+\sqrt2)$.
    3. Differentiate: $\dfrac{dS}{dr}=2\pi r(1+\sqrt2)$.
    4. Approximate error: $dS=2\pi r(1+\sqrt2)\,dr$, with $r=14$, $dr=\dfrac{\sqrt2-1}{11}$.
    5. $dS=2\pi(14)(1+\sqrt2)\cdot\dfrac{\sqrt2-1}{11}=\dfrac{28\pi}{11}\big[(1+\sqrt2)(\sqrt2-1)\big]$.
    6. $(1+\sqrt2)(\sqrt2-1)=(\sqrt2)^2-1^2=2-1=1$ (difference of squares). …

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11. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.If $y=(1+\alpha+\alpha^2+\ldots)e^{nx}$, where $\alpha$ and $n$ are constants, then the relative error in $y$ is (A) error in $x$ (B) percentage error in $x$ (C) $n \cdot (\text{error in } x)$ (D) $n \cdot (\text{Relative error in } x)$

    ›Reveal solutionSolution

    Since the geometric-series prefactor is just a constant, $y$ reduces to $k\,e^{nx}$, and its relative error works out to $n$ times the (absolute) error in $x$.

    Concept and Intuition

    "Relative error in $y$" means $\dfrac{dy}{y}$ (the fractional change), not $dy$ itself. Since $1+\alpha+\alpha^2+\cdots$ doesn't depend on $x$ at all (it's built purely from the constant $\alpha$), it just scales $e^{nx}$ by a fixed factor $k$ — and that scale factor cancels out entirely when we take the ratio $dy/y$, leaving a clean relationship between $y$'s relative error and $x$'s (absolute) error.

    Step-by-Step Solution

    1. Since $|\alpha|<1$ (implicitly, for the series to converge), $1+\alpha+\alpha^2+\cdots = \dfrac{1}{1-\alpha}=k$, a constant.
    2. So $y = k\,e^{nx}$.
    3. Differentiate: $dy = k\cdot n\,e^{nx}\,dx$.
    4. Relative error in $y$ is $\dfrac{dy}{y} = \dfrac{kne^{nx}\,dx}{ke^{nx}} = n\,dx$. …

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12. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
    Q.If the displacement of a particle at time $t$ ($0 < t < \pi$) is given by $s = 3\sin 2t - 6\cos t$, then the acceleration for the values of $t$ at which its velocity is zero is (A) $0$ units$/\mathrm{sec}^2$ (B) $2$ units$/\mathrm{sec}^2$ (C) $3$ units$/\mathrm{sec}^2$ (D) $4$ units$/\mathrm{sec}^2$

    ›Reveal solutionSolution

    This tests differentiating a displacement function twice and correctly restricting the domain when solving the velocity=0 equation. The acceleration at the valid instant is $0$.

    Concept and Intuition

    Velocity is the first time-derivative of displacement, acceleration the second. The subtlety here is that the equation $v=0$ has two algebraic roots, but only one lies in the physically/mathematically allowed range $0<t<\pi$ (where $\sin t\ge 0$), so we must discard the extraneous root before computing acceleration.

    Step-by-Step Solution

    1. $s=3\sin2t-6\cos t$. Differentiate: $v=\dfrac{ds}{dt}=6\cos2t+6\sin t$.
    2. Set $v=0$: $\cos2t+\sin t=0$.
    3. Use $\cos2t=1-2\sin^2t$: $1-2\sin^2t+\sin t=0 \Rightarrow 2\sin^2t-\sin t-1=0$.
    4. Factor: $(2\sin t+1)(\sin t-1)=0 \Rightarrow \sin t=1$ or $\sin t=-\tfrac12$.
    5. For $0<t<\pi$, $\sin t\ge0$ always, so $\sin t=-\tfrac12$ is rejected. Only $\sin t=1$, i.e. $t=\pi/2$, is admissible.
    6. Differentiate $v$ again: $a=\dfrac{dv}{dt}=-12\sin2t+6\cos t$. …

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