NCERT Solutions · Class 12 Mathematics Application of Derivatives

The particle moves on \(\displaystyle 6y=x^{3}+2\), so \(\displaystyle x\) and \(\displaystyle y\) are both functions of time. Differentiating the equation of the curve with respect to \(\displaystyle t\), \[6\,\frac{dy}{dt}=3x^{2}\,\frac{dx}{dt}.\]The condition given is that the \(\displaystyle y\)-coordinate changes \(\displaystyle 8\) times as fast as the \(\displaystyle x\)-coordinate, i.e. \(\displaystyle \dfrac{dy}{dt}=8\,\dfrac{dx}{dt}\). Substituting, \[6\cdot 8\,\frac{dx}{dt}=3x^{2}\,\frac{dx}{dt}\quad\Longrightarrow\quad \left(48-3x^{2}\right)\frac{dx}{dt}=0.\] Since the particle is moving, \(\displaystyle \dfrac{dx}{dt}\neq 0\), so \[3x^{2}=48\quad\Longrightarrow\quad x^{2}=16\quad\Longrightarrow\quad x=4\ \text{or}\ x=-4.\] Both roots must be kept; dropping \(\displaystyle x=-4\) is the usual slip.Now return to the curve for the matching \(\displaystyle y\): \[x=4:\ 6y=4^{3}+2=66\Rightarrow y=11;\qquad x=-4:\ 6y=(-4)^{3}+2=-62\Rightarrow y=-\frac{31}{3}.\]Answer: the required points are \(\displaystyle (4,\,11)\) and \(\displaystyle \left(-4,\,-\dfrac{31}{3}\right)\).

The bubble is a sphere of radius \(\displaystyle r\) cm at time \(\displaystyle t\) seconds, with \(\displaystyle \dfrac{dr}{dt}=\dfrac{1}{2}\ \mathrm{cm}/\mathrm{s}\) and \(\displaystyle V=\dfrac{4}{3}\pi r^{3}\).By the chain rule, \[\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}=4\pi r^{2}\cdot\frac{1}{2}=2\pi r^{2}.\] At \(\displaystyle r=1\ \mathrm{cm}\), \[\frac{dV}{dt}=2\pi(1)^{2}=2\pi.\]Answer: the volume is increasing at \(\displaystyle 2\pi\ \mathrm{cm}^{3}/\mathrm{s}\).

The quantity given is the diameter, so halve it before using the volume formula \(\displaystyle -\) substituting the diameter for \(\displaystyle r\) is the standard error here. \[\text{diameter}=\frac{3}{2}(2x+1)\quad\Longrightarrow\quad r=\frac{3}{4}(2x+1).\]For a sphere \(\displaystyle V=\dfrac{4}{3}\pi r^{3}\), so \[V=\frac{4}{3}\pi\left[\frac{3}{4}(2x+1)\right]^{3}=\frac{4}{3}\pi\cdot\frac{27}{64}(2x+1)^{3}=\frac{9\pi}{16}(2x+1)^{3}.\]Differentiate with respect to \(\displaystyle x\), using the chain rule on \(\displaystyle (2x+1)^{3}\) (the inner derivative is \(\displaystyle 2\)): \[\frac{dV}{dx}=\frac{9\pi}{16}\cdot 3(2x+1)^{2}\cdot 2=\frac{27\pi}{8}(2x+1)^{2}.\]Answer: \(\displaystyle \dfrac{dV}{dx}=\dfrac{27\pi}{8}(2x+1)^{2}\).

Let the cone of sand have base radius \(\displaystyle r\) cm and height \(\displaystyle h\) cm at time \(\displaystyle t\) seconds. The shape condition is \[h=\frac{1}{6}r\quad\Longrightarrow\quad r=6h,\] and it is what lets the two variables be reduced to one. Using it in \(\displaystyle V=\dfrac{1}{3}\pi r^{2}h\) before differentiating, \[V=\frac{1}{3}\pi(6h)^{2}h=\frac{1}{3}\pi\cdot 36h^{3}=12\pi h^{3}.\]Differentiating with respect to \(\displaystyle t\), \[\frac{dV}{dt}=36\pi h^{2}\,\frac{dh}{dt}.\] With \(\displaystyle \dfrac{dV}{dt}=12\ \mathrm{cm}^{3}/\mathrm{s}\), \[\frac{dh}{dt}=\frac{12}{36\pi h^{2}}=\frac{1}{3\pi h^{2}}.\] At \(\displaystyle h=4\ \mathrm{cm}\), \[\frac{dh}{dt}=\frac{1}{3\pi(16)}=\frac{1}{48\pi}.\]Answer: the height is increasing at \(\displaystyle \dfrac{1}{48\pi}\ \mathrm{cm}/\mathrm{s}\).

Marginal cost is defined as the instantaneous rate of change of total cost with respect to the number of units produced, i.e. \(\displaystyle \mathrm{MC}=\dfrac{d\mathrm{C}}{dx}\).With \(\displaystyle \mathrm{C}(x)=0.007x^{3}-0.003x^{2}+15x+4000\), \[\frac{d\mathrm{C}}{dx}=0.021x^{2}-0.006x+15.\]At \(\displaystyle x=17\): \[\frac{d\mathrm{C}}{dx}\bigg|_{x=17}=0.021(289)-0.006(17)+15=6.069-0.102+15=20.967.\]Answer: the marginal cost when \(\displaystyle 17\) units are produced is \(\displaystyle \mathrm{Rs}\ 20.967\).

Marginal revenue is the rate of change of total revenue with respect to the number of units sold, \(\displaystyle \mathrm{MR}=\dfrac{d\mathrm{R}}{dx}\).With \(\displaystyle \mathrm{R}(x)=13x^{2}+26x+15\), \[\frac{d\mathrm{R}}{dx}=26x+26.\]At \(\displaystyle x=7\): \[\frac{d\mathrm{R}}{dx}\bigg|_{x=7}=26(7)+26=182+26=208.\]Answer: the marginal revenue when \(\displaystyle x=7\) is \(\displaystyle \mathrm{Rs}\ 208\).

For a circle, \(\displaystyle A=\pi r^{2}\), so the rate of change of area with respect to the radius is \[\frac{dA}{dr}=2\pi r.\] At \(\displaystyle r=6\ \mathrm{cm}\), \[\frac{dA}{dr}=2\pi(6)=12\pi.\]Answer: option (B), \(\displaystyle 12\pi\).

Marginal revenue is \(\displaystyle \mathrm{MR}=\dfrac{d\mathrm{R}}{dx}\). With \(\displaystyle \mathrm{R}(x)=3x^{2}+36x+5\), \[\frac{d\mathrm{R}}{dx}=6x+36.\] At \(\displaystyle x=15\), \[\frac{d\mathrm{R}}{dx}\bigg|_{x=15}=6(15)+36=90+36=126.\] Note that the answer is the derivative evaluated at \(\displaystyle x=15\), not \(\displaystyle \mathrm{R}(15)\).Answer: option (D), \(\displaystyle 126\).