Related rates: Survey of methods

Problem: We have two quantities, f and g, both depend on time. We want to find some relationship between their derivatives by t.

Solution:
Step 1. Find some relationship between f and g, usually in the form of f = F(g).
Step 2. Differentiate this relationship, using the chain rule on the right.

Example: A glass for wine has the shape of a regular cone with height 8 cm and radius of base equal to 2 cm. We start filling it with wine at the rate 4 cm³ per sec. How fast does the level of wine rise in the glass when it is half full?

Solution: There are two quantities that change, the volume of wine in the glass V and the level of wine h. The cone has parameters H = 8 and R = 2. The rate we were given can be expressed using the variables we just introduced as V ′(t) = 4.

We want to know the rate of change of h. First we need a relationship between V and h, but volume depends on height and radius. To calculate radius based on height we use similarity of triangles.

From this we know that r:h = R:H, therefore r = h/4. Now we easily calculate volume depending on h:

We differentiate by t on both sides, keeping in mind that r is also a function of t. We use the Leibniz notation as it seems better suited to this kind of work.

From this we get

We have an expression for the rate of change of the level of wine, it remains to substitute the given data, in particular h = H/2 = 4. The answer is that when the glass is half full, the level of wine rises at 4/π cm per sec.

For more insight and another example see Related rates in Theory - Applications, see also Solved Problems - Applications.

Global extrema, optimization
Back to Methods Survey - Applications