Question

13-16

(a) What quantities are given in the problem?

(b) What is the unknown?

(c)Draw a picture of the situation for any time \(t\).

(d)Write an equation that relates the quantities.

(e)Finish solving the problem.

14. If a snowball melts so that its surface area decreases at a rate of \(1\,{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{/min}}\), find the rate at which the diameter decreases when the diameter is 10 cm.

Short answer

(a) Given: The surface area decreases with the rate of \(1\,{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{/min}}\). Let \(S\) be surface area then \(\frac{{dS}}{{dt}} = - 1\).

(b) the rate at which the diameter decreases when the diameter is 10 cm.

(c)

(d) \(\frac{{dS}}{{dt}} = 2\pi x\frac{{dx}}{{dt}}\)

(e) \(\frac{{dx}}{{dt}} = - \frac{1}{{20\pi }}\,{\rm{cm/min}}\)

Step-by-step solution

Given quantities (a)

It is given that the surface area decreases with the rate of \(1\,{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{/min}}\).

Suppose \(t\) be the time in minutes and \(S\) be surface area in \({\rm{c}}{{\rm{m}}^{\rm{2}}}\). So, it is given that \(\frac{{dS}}{{dt}} = - 1\).

Write unknown quantities (b)

We need to find the rate at which the diameter decreases when the diameter is 10 cm. Let \(x\) be the diameter. Then the unknown will be \(\frac{{dx}}{{dt}}\) when \(x = 10\,{\rm{cm}}\).

The diagram showing the situation (c)

The diagram showing the situations is given below:

The equation relating the quantities (d)

It is known that the surface area of the sphere is given by \(S = 4\pi {r^2}\). Here, \(r = \frac{x}{2}\). So, the surface area becomes \(S = \pi {x^2}\).

Differentiate the above equation with respect to \(t\) as follows:

\(\begin{aligned}{c}\frac{d}{{dt}}\left( S \right) &= \frac{d}{{dt}}\left( {\pi {x^2}} \right)\\\frac{{dS}}{{dt}} &= \pi \cdot 2x \cdot \frac{{dx}}{{dt}}\\\frac{{dS}}{{dt}} &= 2\pi x \cdot \frac{{dx}}{{dt}}\end{aligned}\)

Hence, \(\frac{{dS}}{{dt}} = 2\pi x \cdot \frac{{dx}}{{dt}}\) shows the relation between the given and required quantities.

Find the unknown quantity (e)

Substitute \(\frac{{dS}}{{dt}} = - 1\), and \(x = 10\) in \(\frac{{dS}}{{dt}} = 2\pi x \cdot \frac{{dx}}{{dt}}\) to find \(\frac{{dx}}{{dt}}\) as follows:

\(\begin{aligned}{c}\frac{{dS}}{{dt}} &= 2\pi x \cdot \frac{{dx}}{{dt}}\\ - 1 &= 2\pi \left( {10} \right) \cdot \frac{{dx}}{{dt}}\\\frac{{dx}}{{dt}} &= - \frac{1}{{20\pi }}\end{aligned}\)

Hence, the rate at which the diameter decreases when the diameter is 10 cm is \(\frac{1}{{20\pi }}\,{\rm{cm/min}}\).