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# Analyzing related rates problems: expressions

When we have a related rates problem on our hands, it's best to first make sure we understand all the involved quantities.

## Want to join the conversation?

• What does A'(t₀) end up being in this problem?
• I think it should be A'(t0) = 8.5 m^2/hour
• When we take the derivative of both sides, wouldn't the derivative of 1/2 = 0 ?
• No. When you take the derivative of both sides, only a constant added onto either side would = 0. If 1/2 was added to the right-hand side of the equation, it would = 0 in the derivative. However, because the 1/2 is a coefficient (and is being multiplied, not added), the 1/2 remains.
This is shown in a derivative rule:
d/dx[A * f(x)] = A * f'(x)
As you can see, the coefficient is kept when the derivative is taken.
• Why is the derivative of the area at that instant "not given"? Khan showed us that we can calculate the derivative at that instant by plugging in.
(1 vote)
• Yes, we can calculate 𝐴'(𝑡₀), but its value is not given in the problem.
• For these problems, is the rate of change that we are looking for always going to be constant? When we solved a problem in the last video, we used measurements from a specific moment to generalize the rate of change of the area of a circle to all times (I think we did this at least).

Can we use the specific quantities at one moment like they usually give us coupled with our given rates of change to make a general rate of change for whatever we're looking for?

Or are there times when, for example, the rate of change of the area of a circle with respect to time is 1 m/s at t=3 and then 3 m/s at t=10?