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.

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.