1. ## Fundamental Question about Related Rates Problems

I am having trouble with this problem:

If y= xˆ2+3 and dx/dt = 2 when x = 1, find dy/dt when x = 1.

The answer is 4 but I thought you could just plug in 1 to the equation, get y = 4, then take the derivative with respect to t of both sides and get dy/dt = 0.

2. ## Re: Fundamental Question about Related Rates Problems

$\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$

Which part are you missing?

3. ## Re: Fundamental Question about Related Rates Problems

well based on the solution in the textbook I think I need dx/dt but I don't understand why.

4. ## Re: Fundamental Question about Related Rates Problems

you have y = f(x) and the problem statement asks for a derivative based in y = g(t). Somehow, you need to get from 'x' to 't'. dx/dt is just the one you need!

Go read that formula I wrote before and ponder it until it soaks in.

5. ## Re: Fundamental Question about Related Rates Problems

I understand why your way 'works,' but could you please explain why I can't simply plug in 1 for x and take the derivative with respect to t on both sides? I guess what I'm saying is my order seems to be different than yours. You take the derivative of each side with respect to t and then you plug in for x, while I do this in reverse. Why can't I do the operations in reverse?

6. ## Re: Fundamental Question about Related Rates Problems

#1 - You have to think about what you are doing. You cannot "plug in" anything, ever! After substitution, you have a constant and the derivative is zero. The entire process you described makes very little sense.

#2 - Take four gears

a) 20 teeth connected via chain to
b1) 10 teeth conneced via welding to
b2) 30 teeth connected via chain to
c) 5 teeth

Now, let's spin a) at 10 revolutions per minute.
This turns b1) at 20 revolutions per minute.
This turns b2) at 20 revolutions per minute.
This turns c) at 120 revolutions per minute.

Ratio of revolutions per minute of a) to b1) is 1:2
Ratio of revolutions per minute of a) to b2) is 1:2
Ratio of revolutions per minute of a) to c) is 1:12
Ratio of revolutions per minute of b1) to b2) is 1:1
Ratio of revolutions per minute of b1) to c) is 1:6
Ratio of revolutions per minute of b2) to c) is 1:6

Your challenge is to get to c) from a) without going through b).
You can also go backwards from c) to a) by going through b).
You cannot go from b) to c) to a).
You cannot go from a) to c) to b).

All things must be in order and in the proper sequence. It "works" because it is in the proper order.

You have y = f(x) and x = g(t). You cannot get to t from y without going through x.

7. ## Re: Fundamental Question about Related Rates Problems

Originally Posted by nicksbyman
I am having trouble with this problem:

If y= xˆ2+3 and dx/dt = 2 when x = 1, find dy/dt when x = 1.

The answer is 4 but I thought you could just plug in 1 to the equation, get y = 4, then take the derivative with respect to t of both sides and get dy/dt = 0.
Since x changes with time and y depends on x, then it should be clear that y will change with time. If you substitute x = 1 (or any other value, for that matter), how can you expect to have y changing with time ....?

8. ## Re: Fundamental Question about Related Rates Problems

Thanks guys I think I get it now. I have thought about it and have realized that if I plug in for x I get y=4, which is a completely different function from what the book asked me to take the derivative of, so I can't do that. Your explanations really helped. Thanks again.

9. ## Re: Fundamental Question about Related Rates Problems

another case for implicit differentiation mentioned in another of your posts ... this time both $x$ and $y$ are implicit functions of time, $t$.

$\frac{d}{dt}(y = x^2 + 3)$

$\frac{dy}{dt} = 2x \cdot \frac{dx}{dt}$

now sub in the the given values for $x$ and $\frac{dx}{dt}$ to determine the rate of change of $y$ w/r to time.