Application of Derivatives Class 12 Maths NCERT Solutions

Step-by-step solution

1. 1Area $A=\pi r^2$, so $\frac{dA}{dr}=2\pi r$.
2. 2(a) At $r=3$: $\frac{dA}{dr}=6\pi$. (b) At $r=4$: $\frac{dA}{dr}=8\pi$.

Step-by-step solution

1. 1Let the edge be $x$. Then $V=x^3$ and $S=6x^2$.
2. 2Differentiate $V=x^3$: $\frac{dV}{dt}=3x^2\frac{dx}{dt}$. At $x=12$: $8=3(144)\frac{dx}{dt}⟹\frac{dx}{dt}=\frac{1}{54}$.
3. 3Differentiate $S=6x^2$: $\frac{dS}{dt}=12x\frac{dx}{dt}=12(12)\left(\frac{1}{54}\right)=\frac{8}{3}$.

Step-by-step solution

1. 1$A=\pi r^2⟹\frac{dA}{dt}=2\pi r\frac{dr}{dt}=2\pi (10)(3)=60\pi$

Step-by-step solution

1. 1$V=x^3⟹\frac{dV}{dt}=3x^2\frac{dx}{dt}=3(100)(3)=900$

Step-by-step solution

1. 1$A=\pi r^2⟹\frac{dA}{dt}=2\pi r\frac{dr}{dt}=2\pi (8)(5)=80\pi$

Step-by-step solution

1. 1$C=2\pi r⟹\frac{dC}{dt}=2\pi \frac{dr}{dt}=2\pi (0.7)=1.4\pi \text{ cm/s}$

Step-by-step solution

1. 1(a) Perimeter $P=2(x+y)$: $\frac{dP}{dt}=2\left(\frac{dx}{dt}+\frac{dy}{dt}\right)=2(-5+4)=-2\text{ cm/min}$.
2. 2(b) Area $A=xy$: $\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}=8(4)+6(-5)=32-30=2{\text{ cm}}^2/\text{min}$.

Step-by-step solution

1. 1$V=\frac{4}{3}\pi r^3⟹\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$.
2. 2At $r=15$ cm: $900=4\pi (225)\frac{dr}{dt}⟹\frac{dr}{dt}=\frac{1}{\pi }\text{ cm/s}$.

Step-by-step solution

1. 1Let the foot be at distance $x$ from the wall and the top at height $y$. Then $x^2+y^2=25$.
2. 2Differentiating: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. At $x=4$ m, $y=3$ m: $2(4)(2)+2(3)\frac{dy}{dt}=0⟹\frac{dy}{dt}=-\frac{8}{3}$.

Step-by-step solution

1. 1Differentiating: $6\frac{dy}{dt}=3x^2\frac{dx}{dt}⟹\frac{dy}{dt}=\frac{x^2}{2}\frac{dx}{dt}$.
2. 2For $\frac{dy}{dt}=2\frac{dx}{dt}$, we need $\frac{x^2}{2}=2⟹x=2$. Then $6y=8+2=10⟹y=\frac{5}{3}$.

Step-by-step solution

1. 1From $\frac{dy}{dt}=\frac{x^2}{2}\frac{dx}{dt}$ and the requirement $\frac{dy}{dt}=8\frac{dx}{dt}$ we get $\frac{x^2}{2}=8⟹x^2=16⟹x=4$.
2. 2Then $6y=64+2=66⟹y=11$.

Step-by-step solution

1. 1$A=\pi r^2⟹\frac{dA}{dt}=2\pi r\frac{dr}{dt}$.
2. 2At $r=6$ cm: $72\pi =2\pi (6)\frac{dr}{dt}⟹\frac{dr}{dt}=6\text{ cm/s}$.

Step-by-step solution

1. 1Let the hypotenuse be $z$ and the two legs $x$, $y$. Then $z^2=x^2+y^2$, with $\frac{dx}{dt}=-2$ and $\frac{dy}{dt}=1$.
2. 2Differentiate: $2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$.
3. 3At $x=18,y=24$ we have $z=30$, so $30\frac{dz}{dt}=18(-2)+24(1)=-12⟹\frac{dz}{dt}=-\frac{2}{5}\text{ cm/s}$.

Step-by-step solution

1. 1Let the man be at distance $x$ from the pole and let his shadow be $s$. By similar triangles: $\frac{s}{x+s}=\frac{2}{6}=\frac{1}{3}$, so $3s=x+s⟹s=\frac{x}{2}$.
2. 2Hence $\frac{ds}{dt}=\frac{1}{2}\frac{dx}{dt}=\frac{1}{2}(5)=2.5\text{ km/h}$.

Step-by-step solution

1. 1If the man is distance $x$ from the lamp and his shadow is $s$, then $\frac{s}{x+s}=\frac{1.5}{4.5}=\frac{1}{3}$, so $3s=x+s⟹s=\frac{x}{2}$.
2. 2Since he walks towards the lamp, $\frac{dx}{dt}=-1$, giving $\frac{ds}{dt}=\frac{1}{2}(-1)=-0.5\text{ m/s}$.

Step-by-step solution

1. 1Differentiate with respect to $t$: $\frac{dy}{dt}=(2x+2)\frac{dx}{dt}$.
2. 2Setting $\frac{dx}{dt}=\frac{dy}{dt}$ gives $2x+2=1⟹x=-\frac{1}{2}$. Then $y=\frac{1}{4}-1=-\frac{3}{4}$.

Step-by-step solution

1. 1Area $A=\frac{1}{2}(12)(7)\sin \theta =42\sin \theta$.
2. 2Then $\frac{dA}{dt}=42\cos \theta \frac{d\theta }{dt}$ with $\frac{d\theta }{dt}=\frac{\pi }{180}\text{ rad/s}$.
3. 3At $\theta =60^{\circ }$: $\frac{dA}{dt}=42\cdot \frac{1}{2}\cdot \frac{\pi }{180}=\frac{7\pi }{60}{\text{ cm}}^2/\text{s}$.

Step-by-step solution

1. 1$f^{\prime }(x)=3>0$ for all $x\in \mathbb{R}$, so $f$ is strictly increasing.

Step-by-step solution

1. 1$f^{\prime }(x)=2e^{2x}>0$ for all $x\in \mathbb{R}$, so $f$ is strictly increasing.

Step-by-step solution

1. 1$f^{\prime }(x)=3x^2-6x+3=3(x-1{)}^2\ge 0$ for all $x$.
2. 2Since $f^{\prime }(x)=0$ only at the isolated point $x=1$, $f$ is strictly increasing on $\mathbb{R}$.

Step-by-step solution

1. 1$f^{\prime }(x)=-\sin x$.
2. 2(a) On $(0,\pi )$ we have $\sin x>0$, so $f^{\prime }(x)<0$: strictly decreasing.
3. 3(b) On $(\pi ,2\pi )$ we have $\sin x<0$, so $f^{\prime }(x)>0$: strictly increasing.

Step-by-step solution

1. 1$f^{\prime }(x)=6x^2-18x+12=6(x-1)(x-2)$.
2. 2Sign of $f^{\prime }(x)$: negative on $(1,2)$, positive on $(-\infty ,1)\cup (2,\infty )$.

Step-by-step solution

1. 1$f^{\prime }(x)=-6x^2+30x-36=-6(x-2)(x-3)$.
2. 2Sign of $f^{\prime }(x)$: negative on $(-\infty ,2)\cup (3,\infty )$, positive on $(2,3)$.

Step-by-step solution

1. 1$f^{\prime }(x)=12+6x-6x^2=-6(x+1)(x-2)$.
2. 2Sign of $f^{\prime }(x)$: negative on $(-\infty ,-1)$ and $(2,\infty )$, positive on $(-1,2)$.

Step-by-step solution

1. 1$f^{\prime }(x)=6(x-1)(x-2)$, same factorisation as above.
2. 2Negative on $(1,2)$, positive on $(-\infty ,1)\cup (2,\infty )$.

Step-by-step solution

1. 1$f^{\prime }(x)=3x^2-12x+9=3(x-1)(x-3)$.
2. 2Negative on $(1,3)$, positive on $(-\infty ,1)\cup (3,\infty )$.

Step-by-step solution

1. 1$f^{\prime }(x)=\frac{1}{x}>0$ for all $x>0$, so $f$ is strictly increasing on $(0,\infty )$.

Step-by-step solution

1. 1$f^{\prime }(x)=2x-1$.
2. 2$f^{\prime }(x)<0$ on $\left(-\infty ,\frac{1}{2}\right)$ and $f^{\prime }(x)>0$ on $\left(\frac{1}{2},\infty \right)$, so the sign changes at $x=\frac{1}{2}$.

Step-by-step solution

1. 1$f^{\prime }(x)=2x+a$. For $f^{\prime }(x)\ge 0$ on $[1,2]$ we need $f^{\prime }(1)=2+a\ge 0$, i.e. $a\ge -2$.

Step-by-step solution

1. 1(i) $(\cos x{)}^{\prime }=-\sin x<0$ on $(0,1)$: decreasing.
2. 2(ii) $(\cos 2x{)}^{\prime }=-2\sin 2x<0$ on $(0,1)$: decreasing.
3. 3(iii) $(\cos 3x{)}^{\prime }=-3\sin 3x<0$ on $(0,1)$: decreasing.
4. 4(iv) $(\tan x{)}^{\prime }={\sec }^{2}x>0$: increasing, not decreasing.

Step-by-step solution

1. 1$f^{\prime }(x)=\frac{1}{\sin x}\cos x=\cot x$.
2. 2On $\left(0,\frac{\pi }{2}\right)$, $\cot x>0$, so $f$ is strictly increasing there.

Step-by-step solution

1. 1$f^{\prime }(x)=\frac{-\sin x}{\cos x}=-\tan x$.
2. 2On $\left(0,\frac{\pi }{2}\right)$, $\tan x>0$, so $f^{\prime }(x)<0$ and $f$ is strictly decreasing.

Step-by-step solution

1. 1$y^{\prime }=\frac{1}{1+x}-\frac{(2)(2+x)-2x(1)}{(2+x{)}^2}=\frac{1}{1+x}-\frac{4}{(2+x{)}^2}$.
2. 2So $y^{\prime }=\frac{(2+x{)}^2-4(1+x)}{(1+x)(2+x{)}^2}=\frac{x^2}{(1+x)(2+x{)}^2}\ge 0$ for all $x>-1$.
3. 3Since $y^{\prime }\ge 0$ with equality only at $x=0$, $y$ is increasing on its domain.

Step-by-step solution

1. 1$f^{\prime }(x)=\cos x-\sin x=\sqrt{2}\cos \left(x+\frac{\pi }{4}\right)$.
2. 2$f^{\prime }(x)\ge 0$ when $\cos \left(x+\frac{\pi }{4}\right)\ge 0$, i.e. $0\le x\le \frac{3\pi }{4}$ or $\frac{7\pi }{4}\le x\le 2\pi$.
3. 3Hence strictly increasing on $\left(0,\frac{3\pi }{4}\right)$ and $\left(\frac{7\pi }{4},2\pi \right)$; strictly decreasing on $\left(\frac{3\pi }{4},\frac{7\pi }{4}\right)$.

Step-by-step solution

1. 1$f^{\prime }(x)=3x^2-6x+3=3(x-1{)}^2\ge 0$ for all $x$, equality only at $x=1$.
2. 2So $f$ is increasing on $\mathbb{R}$.

Step-by-step solution

1. 1$y^{\prime }=3x^2-1$.
2. 2At $x=2$: $y^{\prime }=3(4)-1=11$.

Step-by-step solution

1. 1Tangent slope at $x=2$ is 11, so the normal slope is its negative reciprocal: $-\frac{1}{11}$.

Step-by-step solution

1. 1$y^{\prime }=3x^2-3$.
2. 2At $x=3$: $y^{\prime }=27-3=24$.

Step-by-step solution

1. 1$y^{\prime }=3x^2-3$.
2. 2At $x=2$: $y^{\prime }=12-3=9$.

Step-by-step solution

1. 1$y^{\prime }=3x^2-3$.
2. 2At $x=1$: $y^{\prime }=3-3=0$.

Step-by-step solution

1. 1At $x=0$, $y=5$, so the point is $(0,5)$.
2. 2$y^{\prime }=4x^3-18x^2+26x-10$, so at $x=0$, $y^{\prime }=-10$.
3. 3Tangent: $y-5=-10(x-0)⟹y=-10x+5$.

Step-by-step solution

1. 1Tangent slope at $(0,5)$ is $y^{\prime }(0)=-10$, so normal slope is $\frac{1}{10}$.

Step-by-step solution

1. 1$y^{\prime }=4x^3-18x^2+26x-10$, so at $x=2$, $y^{\prime }=32-72+52-10=2$.
2. 2Normal slope $=-\frac{1}{y^{\prime }}=-\frac{1}{2}$.

Step-by-step solution

1. 1$y^{\prime }=\frac{(1)(x-2)-(x-1)(1)}{(x-2{)}^2}=\frac{-1}{(x-2{)}^2}$.
2. 2At $x=3$: $y^{\prime }=-1$.

Step-by-step solution

1. 1$y^{\prime }=\frac{-1}{(x-2{)}^2}$.
2. 2At $x=4$: $y^{\prime }=-\frac{1}{4}$.

Step-by-step solution

1. 1$\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{3a{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta }=-\tan \theta$.
2. 2At $\theta =\frac{\pi }{4}$: $\frac{dy}{dx}=-1$, so normal slope $=1$.

Step-by-step solution

1. 1$\frac{dy}{dx}=-\tan \theta$, which is undefined (vertical tangent) at $\theta =\frac{\pi }{2}$.
2. 2The tangent is vertical, hence the normal is horizontal: slope $=0$.

Step-by-step solution

1. 1Slope of chord $=\frac{2-(-2)}{2-1}=4$.
2. 2Set $y^{\prime }=3x^2-3=4⟹x^2=\frac{7}{3}⟹x=\pm \sqrt{\frac{7}{3}}$.
3. 3The points are $\left(\pm \sqrt{\frac{7}{3}},\pm \sqrt{\frac{7}{3}}\left(\frac{7}{3}-3\right)\right)$.

Step-by-step solution

1. 1$y^{\prime }=2x-2$.
2. 2(a) Slope of the given line is 2, so $2x-2=2⟹x=2$. Then $y=4-4+7=7$. Tangent: $y-7=2(x-2)$.
3. 3(b) Slope of $5y-15x=13$ is $3$, so needed tangent slope is $-\frac{1}{3}$: $2x-2=-\frac{1}{3}⟹x=\frac{5}{6}$. Then $y=\frac{25}{36}-\frac{10}{6}+7=\frac{217}{36}$.

Step-by-step solution

1. 1$y^{\prime }=3x^2-4x+1$, so at $x=1$: $y^{\prime }=3-4+1=0$.
2. 2Tangent: $y+1=0(x-1)⟹y=-1$.

Step-by-step solution

1. 1Differentiate $a{y}^2=x^3$: $2ay{y}^{\prime }=3x^2⟹y^{\prime }=\frac{3x^2}{2ay}$.
2. 2At $(a{m}^2,a{m}^3)$: $y^{\prime }=\frac{3a^2{m}^4}{2a^2{m}^3}=\frac{3m}{2}$.
3. 3Normal slope $=-\frac{2}{3m}$. Normal: $y-a{m}^3=-\frac{2}{3m}\left(x-a{m}^2\right)$.

Step-by-step solution

1. 1$y^{\prime }=3x^2-6x+2$.
2. 2At $x=0$: $y^{\prime }=2$.
3. 3At $x=1$: $y^{\prime }=3-6+2=-1$.

Step-by-step solution

1. 1Parallel to the $x$-axis means $y^{\prime }=12x^2-4x-9=0$.
2. 2$x=\frac{4\pm \sqrt{16+432}}{24}=\frac{4\pm \sqrt{448}}{24}=\frac{1\pm \sqrt{7}}{6}$.

Step-by-step solution

1. 1Differentiate: $2x+2y{y}^{\prime }=0⟹y^{\prime }=-\frac{x}{y}$.
2. 2At $(4,3)$: $y^{\prime }=-\frac{4}{3}$.
3. 3Tangent: $y-3=-\frac{4}{3}(x-4)⟹4x+3y=25$.