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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 9

Lesson 5: Solving motion problems using parametric and vector-valued functions- Planar motion example: acceleration vector
- Planar motion (differential calc)
- Motion along a curve: finding rate of change
- Motion along a curve: finding velocity magnitude
- Motion along a curve (differential calc)
- Planar motion (with integrals)
- Planar motion (with integrals)

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# Motion along a curve: finding rate of change

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.

## Want to join the conversation?

- 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?(5 votes)
- 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. :)(5 votes)
- 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.(4 votes)
- Why is the dx/dt and dy/dt have t in it. Is it because where mesuring time?(2 votes)
- yes, absolutely correct, we are measuring the change with respect to Time.(4 votes)

- 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.(3 votes) - 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.(2 votes)
- yes, but you still would have had to do the quotient rule... which is essentially a combination of the product & chain rule. :P(3 votes)

- At2:51, why is d/dt [ x^2 ] equal to 2x * dx/dt? Should it not be 2x* d(x^2)/dt?(2 votes)
- 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!(2 votes)

- At2:43doesn'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.(3 votes)

- 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:)(0 votes)- In math there is never
**one**way to solve a problem, though that doesn't mean that all methods are equally simple. Your method works and makes sense, though the fact it is more complex means that there is more room for error, as long as you check yourself you'll be good though.

Note: There is a hickup in it for when you got the individual function for y(t) where you forgot to put ±4/x when taking the square root.(1 vote)

## Video transcript

- [Voiceover] We're told that a particle moves along the curve x squared, y squared is equal to 16, so that
the the x-coordinate is changing at a constant rate of negative two units per minute. What is the rate of
change in units per minute of the particle's y-coordinate when the particle is at the point 1, 4? So let's just repeat, or
rewrite, what they told us. So the curve is described by x squared, y squared is equal to 16. They tell us that up there. They tell us that the x-coordinate is changing at a constant rate. We underline that. The x-coordinate is changing at a constant rate of negative
two units per minute. So we could say that dx ... I'll write over here
on the right hand side. Dx, dt. The rate of change of the x-coordinate with respect to time is
equal to negative two. And they're saying units,
some unit of distance, units divided by minute. Units per minute. What they want us to figure out is what is the rate of change of the particle's y-coordinate. So let me underline that. What is the rate of change of
the particle's y-coordinate? So what they want us to
find is what is dy, dt. What is that equal to? And they say when the
particle is at the point 1,4, so when x is equal to
one, so x is equal to one. And y is equal to four,
y is equal to four. So can we set up some
equation that involves the rate of change of x with respect to t, y with respect to t, x and y? Well what if we were
to take the derivative of this relation that describes the curve? What if we were to take the derivative with respect to t on both sides? So let me write that down. So we're going to take the derivative. Actually let me just erase this so I have a little bit more space. Alright. That way I can just add it. So let's take the derivative with respect to t on both sides of that. And if at any point you get inspired I encourage you to pause the video and try to work through it. Well on the left hand side, if we view this as a product of two
functions right over here, we could take the derivative
of the first function, which is going to be the derivative of x squared with respect to x. So that is 2x, and remember we're not just taking the derivative
with respect to x, we're taking the derivative
with respect to t, so we're going to have
to apply the chain rule. So it's going to be the
derivative of x squared with respect to x, which is 2x, times the derivative
of x with respect to t. So times dx, dt. And then we're going to multiply that times the second function. So times y squared. Times y squared. And then that's going to be plus the first function,
which is just x squared times the derivative of the second function with respect to d. So once again we're going
to apply the chain rule. The derivative of y-squared
with respect to y is 2y. I'm gonna do that in that orange color. It is equal to 2y. And then times the derivative
of y with respect to t. Times dy, dt. And then that is going to be equal to the derivative with respect to t of 16. Well that doesn't change over time, so that's just going to be equal to zero. So here we have it. We need to simplify this a little bit. But we have an equation that gives our relationship between
x, derivative of x with respect to t, y, and
derivative of y with respect to t. So actually let me just rewrite it one more time so it's
a little bit simplified. So this is 2xy squared dx, dt, plus ... Actually I don't even
have to rewrite it again. All we're trying to do
is solve for dy, dt, so let's actually just
substitute the values in. So we know we want to figure out what's going on when x is equal to one. So we know that the x's
here are equal to one. This x, x squared, well that's
just going to be one squared, so that's just going to be equal to one. We know that y is equal to four. So this is going to be 16 and
this is going to be eight. We know the derivative of x with respect to t is negative two. They tell us that in
the problem statement. Negative two. So now this is a good time
to simplify this thing. So this will simplify to ... Let's see, all of this is going to be two times one times negative two, so that is negative four times 16. So that is negative 64,
and then we have ... We'll use a color you can see. And then we have all of this. Well this is just going to be
one time eight times dy, dt. So this going to be eight dy, dt. So plus eight times the derivative of y with respect to t is equal to zero. Add 64 to both sides and we get ... I'll switch to a neutral color. Eight times the derivative of y with respect to t is equal to 64. Divide both sides by eight and you get the derivative of y with respect to t is equal to 64 divided by eight is just 8. If you want to look at the units it will also be in units per minute. Some units of distance per minute. And we are done.