# Related rates

• 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