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# Motion along a curve: finding rate of change

AP.CALC:
FUN‑8 (EU)
,
FUN‑8.B (LO)
,
FUN‑8.B.1 (EK)
Given that a particle moves along the implicit curve x²y²=16 and given the rate of change of x with respect to t at some point, Sal finds the particle's rate of change with respect to y using implicit differentiation.

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• Ok, so each question in this skill is completely different from the last. Is there a one formula that you use for each problem that you may slightly change or what? i cant solve this problem, i've watched this video and and looked at all the hints but it still makes no sense. Can someone simplify it for me or use an analogy if possible for me?
• I believe there is a mistake in the third line:

d/dx [f(g(x))] = df/dg * dg/dx = df/dx
If we change some letters we get:
d/dx [y(x(t))] = dy/dx * dx/dt = dy/dt

I believe first term in the third line should be d/dt.

This was the cause of a little confusion for me...
(1 vote)
• You can also solve the original equation for y to get y=4/x, then find dy/dt = (-4/x^2)*dx/dt. Then plug in dx/dt=-2, and x=1 to get dy/dt=(-4)(-2) = 8. :)
• Why is the dx/dt and dy/dt have t in it. Is it because where mesuring time?
• yes, absolutely correct, we are measuring the change with respect to Time.
• I was curious about the effect of keeping 16 on the RHS all by its lonesome, so I moved the x^2 to the RHS so the 16 was not reduced to 0 by taking the derivative. The answer, thankfully was the same. I think my was simpler, no product rule required.
• yes, but you still would have had to do the quotient rule... which is essentially a combination of the product & chain rule. :P
• I think Sal's approach is more robust than the one I took. I decided to derive parameterized equations of positions of x and y. Since the x-coordinate has a constant velocity of -2, it made sense to me that its position in terms of t can be defined as -2t. I then recast the original equation in terms if x and y and got y = 4/x. Substituting -2t for x, I got y = -2/t, so the position vector is (-2t, -2/t). Then I derived the velocity vector as (-2, 2t^-2). Next, I needed to know at what t the particle would be at the point (1,4), so computed that from 4 = 2/t, which resulted in t = -1/2. I then plugged that into the velocity equation for y (i.e. 2(-1/2)^-2), which resulted in 8.

I think the insight that the x position can be described as -2t is easy to do in this specific problem, but that would not be the case in more complex ones.
• At doesn't he mean that we need to use the "product rule" instead of the chain rule?
(1 vote)
• in order to get the derivative since it was x^2 and y^2, you need to apply not just the product rule when multiplying one times the other, but also the chain rule to get the derivative of x^2 and y^2 themselves.
• At , why is d/dt [ x^2 ] equal to 2x * dx/dt? Should it not be 2x* d(x^2)/dt?
(1 vote)
• Taking `d/dt[x^2]` requires using the chain rule. If we let `f(x) = x^2`, then we can expand `d/dt[f]` as

`df/dt = df/dx * dx/dt`.

Looking at the final expression, `df/dx * dx/dt`, we can verify that the equality indeed holds true because the `dx` terms cancel (leaving us with the original `df/dt`). From here, it's a matter of using power rule to find `df/dx`:

`df/dx = d/dx[f] = d/dx[x^2] = 2x`

Then, looking back at the equality that we already found, `df/dt = df/dx * dx/dt`, we can just substitute the `df/dx` with `2x` to simplify the expression:

`df/dt = df/dx * dx/dt = 2x * dx/dt`

If any of this is unfamiliar, I recommend you search for the chain rule videos on Khan Academy. Hope this helps!
• Alternatively, dy/dx=(dy/dt)/(dx/dt) therefore dy/dt=dy/dx*dx/dt. At least for this problem, you only need to implicitly differentiate y in respect to x then multiply by dx/dt (which was equal to -2). I'm not sure if this would work in other circumstances, but at least when they give you a defined value for dy/dt or dx/dt it seems like this will work.
(1 vote)
• For any y=f(x) function, the derivative (rate of change) of y assumes that the rate of change of x is 1.

So if you tell me that the x-coordinate's rate of change is some multiple of that, then to find the derivative of y, I'll just calculate the derivative using the familiar methods taught here, and then multiply that derivative by whatever the x-coordinate's rate of change is.

This is not the calculation method modeled in Sal's videos here. Is there some pitfall to using my approach? (I haven't arrived at a wrong answer yet!)
(1 vote)
• I'm not too sure of what you mean, but if the method works, great! Try to see if it works for all cases and maybe even figure out why it works.
Here is my opinion on it. In this problem, y is not explicitly defined as a function of x, so implicit differentiation is used. Your statement of "For any y=f(x) function, the derivative (rate of change) of y assumes that the rate of change of x is 1." is a little confusing for me, but I assume you meant that the rate of change of x with respect to x is 1. However, in this problem we differentiate with respect to t, where dx/dt = -2. I believe that with more difficult problems, implicit differentiation should be used, as it may be harder to see the reasoning as to where the "multiply that derivative by whatever the x-coordinate's rate of change is" comes from.
Hope that I helped, correct me if I'm wrong.
(1 vote)
• When I paused the video to try solving the problem before the explanation I used a different method. My question is would this always work or should I stick to the method Sal used?
This is the method I used:
to find the derivative of a parametric function we use this formula: dy/dx = (dy/dt)/(dx/dt).
The curve can be re-written as y = 4/x, so dy/dx is -4/x^2.
Substituting we find: -4/x^2 = (dy/dt)/-2
We're dealing with the point (1, 4) so: -4 = (dy/dt)/-2
Solve for dy/dt: dy/dt = 8.

Could it be that this doesn't work if for each x-value there are more y values?? Thank you in advance for the answer:)