Example 1: Differential Equation Application

The rate at which a snowball rolling down a hill gains weight is proportional to the difference between its weight at the bottom of the hill and its current weight. At time t=0 , when the snowball is weighed at the top of the hill, its weight is 10 grams. If $S\left(t\right)$ is the weight of the snowball, in grams, at time t seconds after it is first weighed, then

$\frac{\mathit{dS}}{\mathit{dt}}=\frac{1}{10}(100–S)$.

Let $y=S\left(t\right)$ be the solution to the differential equation above with initial condition $S\mathrm{(}0\mathrm{)}=10$.

Is the snowball gaining weight faster when it weighs 14 grams or when it weighs 17 grams?

Identifier: You are given to a differential equation to work with, and being asked a question about how a rate is behaving.

Step 1: Plug your given snowball weights, 14 grams and 17 grams , into the given differential equation.

$\frac{\mathit{dS}}{\mathit{dt}}=\frac{1}{10}(100–S)$

Notice that the independent variable (what you need to plug into the differential equation) is the S-value , which is the weight of the snowball.

The given differential equation, $\frac{\mathit{dS}}{\mathit{dt}}$, provides you information about the change in the S-variable (snowball weight in grams) per the change in the t-variable (seconds) .

$\frac{\mathit{dS}}{\mathit{dt}}=\frac{\mathit{change\; in\; snowball\; weight}}{\mathit{change\; in\; time}}=\frac{\mathit{gram}}{\mathit{second}}$

$\frac{\mathit{dS}}{\mathit{dt}}=\frac{1}{10}(100–S)$

${\frac{\mathit{dS}}{\mathit{dt}}|}_{\mathrm{S}=14}=\frac{1}{10}\mathrm{(}100–14\mathrm{)}=8.6\frac{\mathit{gram}}{\mathit{second}}$

${\frac{\mathit{dS}}{\mathit{dt}}|}_{\mathrm{S}=17}=\frac{1}{10}\mathrm{(}100–17\mathrm{)}=8.3\frac{\mathit{gram}}{\mathit{second}}$

Step 2: The question wants you to find which snowball was gaining weight faster. In other words, which one has the greatest amount of $\frac{\mathit{gram}}{\mathit{second}}$.

To find that value, compare your differential equation results from Step 1.

${\frac{\mathit{dS}}{\mathit{dt}}|}_{\mathrm{S}=14}=8.6\frac{\mathit{gram}}{\mathit{second}}$

${\frac{\mathit{dS}}{\mathit{dt}}|}_{\mathrm{S}=17}=8.3\frac{\mathit{gram}}{\mathit{second}}$

$8.6\frac{\mathit{gram}}{\mathit{second}}>8.3\frac{\mathit{gram}}{\mathit{second}}$

${\frac{\mathit{dS}}{\mathit{dt}}|}_{\mathrm{S}=14}>{\frac{\mathit{dS}}{\mathit{dt}}|}_{\mathrm{S}=17}$

Final Result:

The snowball is gaining weight at the rate of $8.6\frac{\mathit{gram}}{\mathit{second}}$ when it weighs 14 grams , and it is gaining weight at the rate of$8.3\frac{\mathit{gram}}{\mathit{second}}$ when it weighs 17 grams . The snowball weight is therefore growing faster (larger rate of change ) when it weighs 14 grams since it is adding weight at the rate of 8.6grams per second.