Question

A cylinder gets taller at a rate of \[3\] inches per second, but the radius shrinks at a rate of \[1\] inch per second. How fast is the volume of the cylinder changing when the height is \[20\] inches and the radius is \[10\] inches?

Answer 1

Hint: Given that the cylinder gets taller at a rate of \[3\] inches per second i.e., \[\dfrac{{dh}}{{dt}} = 3\] inches per second and also given, the radius shrinks at a rate of \[1\] inch per second i.e., \[\dfrac{{dr}}{{dt}} = - 1\] inch per second. And we know that the volume of a cylinder is \[V = \pi {r^2}h\], where \[r\] is the radius of the cylinder and \[h\] is the height of the cylinder. We will differentiate \[V\] w.r.t to \[t\] and then we will substitute the values of \[\dfrac{{dh}}{{dt}}\] and \[\dfrac{{dr}}{{dt}}\]. We will calculate the required rate of change of the volume by putting \[h\] as \[20\] inches and \[r\] as \[10\] inches.

Complete step by step answer:
Given that the cylinder gets taller at a rate of \[3\] inches per second i.e., \[\dfrac{{dh}}{{dt}} = 3\] inches per second and also given, the radius shrinks at a rate of \[1\] inch per second i.e., \[\dfrac{{dr}}{{dt}} = - 1\] inch per second. Now, as we know that volume of a cylinder is given by,
\[\pi {r^2}h\]
where \[r\] is the radius of the cylinder and \[h\] is the height of the cylinder i.e., \[V = \pi {r^2}h\].

Differentiating both the sides of \[V = \pi {r^2}h\] with respect to \[t\], we get
\[ \Rightarrow \dfrac{d}{{dt}}\left( V \right) = \dfrac{d}{{dt}}\left( {\pi {r^2}h} \right)\]
Since, \[\pi \] is a constant. So, we get
\[ \Rightarrow \dfrac{{dV}}{{dt}} = \pi \dfrac{d}{{dt}}\left( {{r^2}h} \right)\]
From the product rule of differentiation, we know \[\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]. Using this, we get
\[ \Rightarrow \dfrac{{dV}}{{dt}} = \pi \left[ {{r^2}\dfrac{d}{{dt}}\left( h \right) + h\dfrac{d}{{dt}}\left( {{r^2}} \right)} \right]\]

On differentiating, we get
\[ \Rightarrow \dfrac{{dV}}{{dt}} = \pi \left[ {{r^2}\dfrac{{dh}}{{dt}} + \left( {2r} \right)h\dfrac{{dr}}{{dt}}} \right]\]
On rewriting, we get
\[ \Rightarrow \dfrac{{dV}}{{dt}} = \pi {r^2}\dfrac{{dh}}{{dt}} + 2\pi rh\dfrac{{dr}}{{dt}}\]
Using the given values of \[\dfrac{{dh}}{{dt}}\] and \[\dfrac{{dr}}{{dt}}\], we get
\[ \Rightarrow \dfrac{{dV}}{{dt}} = \pi {r^2} \times \left( 3 \right) + 2\pi rh \times \left( { - 1} \right)\]

Now, we have to find how fast the volume of the cylinder changes when the height is \[20\] inches and the radius is \[10\] inches. Therefore, we get
\[ \Rightarrow \dfrac{{dV}}{{dt}} = \pi \times {\left( {10} \right)^2} \times \left( 3 \right) + 2\pi \times \left( {10} \right) \times \left( {20} \right) \times \left( { - 1} \right)\]
On simplifying, we get
\[ \Rightarrow \dfrac{{dV}}{{dt}} = 300\pi - 400\pi \]
\[ \therefore \dfrac{{dV}}{{dt}} = 100\pi \]

Therefore, the volume of the cylinder changes at a rate of \[100\pi {\text{ }}\dfrac{{inche{s^3}}}{s}\] when the height is \[20\] inches and the radius is \[10\] inches.

Note: When a value \[y\] varies with \[x\] such that it satisfies \[y = f(x)\], then the rate of change of \[y\]with respect to \[x\] is \[f'(x) = \dfrac{{dy}}{{dx}}\]. If two variable \[x\] and \[y\] vary with respect to another variable \[t\] such that \[x = f(t)\] and \[y = g(t)\], then using chain rule we can define \[\dfrac{{dy}}{{dx}} = \dfrac{{\left( {\dfrac{{dy}}{{dt}}} \right)}}{{\left( {\dfrac{{dx}}{{dt}}} \right)}}\], if \[\dfrac{{dx}}{{dt}} \ne 0\].