- In the problem type of
*related rates*, based on a verbal description, you have to set up an equation, which you then have to derivate implicitly to get the solution.
- Example. 3.9.2.
- Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 1 m/s, how fast is the area of the spill increasing when the radius is 30 m?
- You can see that this is a related rates problem from that it's asking about the
*rate of change* of the area of the spill.
- Since the rate of change of the radius is given in m/s, you can see that the variable with respect to which you'll have to derivate is the time, measured in seconds.
- Let's set up notation: time is \(t\), radius is \(r\), area is \(A\).
- This problem type is called related rates, because the rate of change of \(r\) is related to the rate of change of \(A\).
- We have to find the relation between \(r\) and \(A\).
- It's the area formula for a circular disk: \[
A=r^2\pi
\]
- Now we have to use implicit differentiation on this, with respect to \(t\): \[
A'=2rr'\pi
\]
- Now we can substitute the numerical values given in the problem: \[
A'=2\cdot30\cdot1\cdot\pi=60\pi\,\mathrm m/\mathrm s.
\]

## Example 2

- 3.9.12.
- A particle is moving along a hyperbola \(xy=8\). As it reaches the point \((4,2)\), the \(y\)-coordinate is decreasing at a rate of 3 cm/s. How fast is the \(x\)-coordinate of the point changing at that instant?
- Here, the expression relating \(x\) and \(y\) is already given to us, so we'll have to derivate it implicitly.
- What you need to look out for is that the variable with respect to which we're differentiating is \(t\), the time.
- Therefore implicit differentiation yields the following. \[
x'y+xy'=0
\]
- Substituting the known numerical values, we get the following. \[
x'\cdot2+4\cdot(-3)=0.
\]
- This yields \(x'=6\) cm/s.

## Example 3

- 3.9.14.
- If a snowball melts so that its surface area decreases at a rate of \(1\,\mathrm{cm}^2/\mathrm{min}\), find the rate at which the diameter decreases when the diameter is 10 cm.
- The area of a sphere is \(A=4r^2\pi\). But we need an equation with the area \(A\), and the diameter \(d=2r\).
- Therefore, we want to use \[
A=4(d/2)^2\pi=d^2\pi.
\]
- Implicit differentiation gives \[
A'=2dd'\pi
\]
- Substituting the given numerical values, we get \[
-1=2\cdot10\cdot d'\cdot\pi
\]
- Therefore, we get \(d'=-\frac{1}{20\pi}\) cm/min.

- Exercises. 3.9: 4, 6, 8, 16, 18, 22, 24, 30, 38