# The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball. Suppose the snowball is perfectly spherical. Then the volume (in centimeters cubed) of a ball of radius r centimeters is 4/3 pi r^3. The surface area is 4 pi r^2.Set up the differential equation for how r is changing. Then, suppose that at time t = 0 minutes, the radius is 10 centimeters. After 5 minutes, the radius is 8 centimeters. At what time t will the snowball be completely melted.

Question
Upper Level Math
The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball. Suppose the snowball is perfectly spherical. Then the volume (in centimeters cubed) of a ball of radius r centimeters is $$\displaystyle\frac{{4}}{{3}}\pi{r}^{{3}}$$. The surface area is $$\displaystyle{4}\pi{r}^{{2}}$$.Set up the differential equation for how r is changing. Then, suppose that at time t = 0 minutes, the radius is 10 centimeters. After 5 minutes, the radius is 8 centimeters. At what time t will the snowball be completely melted.

2021-01-03
Step 1
Given:
The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball is perfectly spherical
Then the volume (in centimetres cubed) of a ball of radius r centimetres is
$$\displaystyle{v}=\frac{{4}}{{3}}\pi{r}^{{3}}$$
And the surface area is
$$\displaystyle{s}={4}\pi{r}^{{2}}$$
Set up the differential equation for how r is changing.
Then, suppose that at time t=0 minutes,the radius is 10 centimetres.After 5 minutes,the radius is 8 centimetres
Step 2
To find: At the what time t will be snowball be completely melted?
From the given conditions ,
$$\displaystyle\frac{{{d}{V}}}{{{\left.{d}{t}\right.}}}{\left(\alpha{S}\right)}$$
$$\displaystyle\frac{{{d}{V}}}{{{\left.{d}{t}\right.}}}{\left(\lambda{S}\right)}$$......(1)
$$\displaystyle{V}=\frac{{4}}{{3}}\pi{r}^{{3}}$$
$$\displaystyle\frac{{{d}{V}}}{{{\left.{d}{t}\right.}}}=\frac{{4}}{{3}}\pi{3}{r}^{{2}}\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}$$
By putting this value in (1)
The equation must be,
$$\displaystyle\frac{{4}}{{3}}\pi{3}{r}^{{2}}\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}=\lambda{4}\pi{r}^{{2}}$$
$$\displaystyle\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}=\lambda$$
$$\displaystyle{r}=\lambda{t}+{c}$$
Now, t = 0 and r = 10
so,
$$\displaystyle{r}=\lambda{t}+{c}$$
$$\displaystyle{10}=\lambda{\left({0}\right)}+{c}$$
c=10 and here
$$\displaystyle{r}=\lambda{t}+{10}$$......(2)
After the 5 minutes, t = 5 and r = 8
$$\displaystyle{8}=\lambda{\left({5}\right)}+{10}$$
$$\displaystyle{5}\lambda=-{2}$$
$$\displaystyle\lambda=-\frac{{2}}{{5}}$$
Now equation (2)becomes,
$$\displaystyle{r}=-\frac{{2}}{{5}}{t}+{10}$$......(3)
This shows the differential equation for how r is changing.
As the snowball completely melted that means the radius of the snowball is zero.
From this by substituting r=0 in the equation (3) must be,
$$\displaystyle{0}=-\frac{{2}}{{5}}{t}+{10}$$
$$\displaystyle\frac{{2}}{{5}}{t}={10}$$
$$\displaystyle{t}=\frac{{{10}\times{5}}}{{2}}$$ t=25 Hence the time required for melting the snowball is 25 minutes.

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