 Burhan Hopper

2021-01-02

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 $\frac{4}{3}\pi {r}^{3}$. 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. 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
$v=\frac{4}{3}\pi {r}^{3}$
And the surface area is
$s=4\pi {r}^{2}$
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 ,
$\frac{dV}{dt}\left(\alpha S\right)$
$\frac{dV}{dt}\left(\lambda S\right)$......(1)
$V=\frac{4}{3}\pi {r}^{3}$
$\frac{dV}{dt}=\frac{4}{3}\pi 3{r}^{2}\frac{dr}{dt}$
By putting this value in (1)
The equation must be,
$\frac{4}{3}\pi 3{r}^{2}\frac{dr}{dt}=\lambda 4\pi {r}^{2}$
$\frac{dr}{dt}=\lambda$
$r=\lambda t+c$
Now, $t=0$ and $r=10$
so,
$r=\lambda t+c$
$10=\lambda \left(0\right)+c$
c=10 and here
$r=\lambda t+10$......(2)
After the 5 minutes, $t=5$ and $r=8$
$8=\lambda \left(5\right)+10$
$5\lambda =-2$
$\lambda =-\frac{2}{5}$
Now equation (2)becomes,
$r=-\frac{2}{5}t+10$......(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,
$0=-\frac{2}{5}t+10$
$\frac{2}{5}t=10$
$t=\frac{10×5}{2}$ $t=25$ Hence the time required for melting the snowball is 25 minutes.

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