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# Motion along a curve: finding velocity magnitude

AP.CALC:
FUN‑8 (EU)
,
FUN‑8.B (LO)
,
FUN‑8.B.1 (EK)
Given that a particle moves along the implicit curve xy=16 and given its rate of change with respect to y at some point, Sal finds the magnitude of the particle's velocity vector using implicit differentiation.

## Want to join the conversation?

• I am still really confused about doing d/dx or dy/dx or dx/dt and so on and so forth.

I know that "d/dx" = "take the derivative of everything in the equation that has an 'x'"

but if it doesn't have an 'x' you just place a "d(whatever)/dx"? is that correct? and then somehow later find a way to find that "d(whatever)/dx"?

For example. I am confused where t comes in this equation. t was never given in the original equation, so how can you just add it in?

Thank you.

2nd half of question!

okay how is

d/dt of x = 1*dx/dt?

Why isn't it just one? or dx/dt? in other words, why do you use the chain rule for doing a d/dt of x?
• This website uses notation that differs from my calculus textbook, which also differs (on occasion) from the notation in my calculus class! There are lots of derivative notations floating around. It is confusing and a video explaining them all might be helpful. I'll try my best here.

Jelle, you are correct, the notation "d/dx" can also be stated as "the derivative with respect to x." We can also write it as "dy/dx," aka "the derivative of y (the function) with respect to x." Of course, we can name a function pretty much anything other than y, hence notations such as "db/dt" or "dr/dt" and the list goes on (you'll use such notation in related rate problems). Notation "d/dt" or "dx/dt" or even "dy/dt" can be stated as "the derivative (in general, or of x or y) with respect to t," with t being a third parameter. Again, regarding related rate problems, or acceleration or velocity problems, t usually stands for time, but it can also stand for some third parameter when differentiating.

To clarify (x)(dx/dt) = 1(dx/dt), we treat it similar to the way we treat y in implicit differentiation: We take the derivative of x, which is 1, and multiply it by x', similar to how 2y(dy/dt) = 2(dy/dt) or 2y'. Since we are differentiating every variable in such equations with respect to a third parameter, this explains why every variable has its own derivative notation.

To try and sum the ones I know:
f'(x) = [letter representing function name, usually y]' = d[letter representing function name, again usually y]/dx (or dt, if a third parameter, such as time, applies). Of course the symbol ' is read as "prime."

More explicitly: f'(x) = y' = dy/dx or dy/dt.

I hope this is helpful and accurate. :)
• Wish he'd graph the thing so I could understand what's actually going on
• If you can use Google to do searches, you can do this yourself!

Just enter `16/x`

Of course that won't show you motion over time ...
• Beginning at , Mr. Khan is trying to explain how to solve for dx/dt. He states: "We're going to use a little bit of the product rule and a little bit of the chain rule here." Then he proceeds to apply the product rule, which is justified by the nature of the left hand side of the equation x·y, but he never uses the chain rule for the obvious reason that it's inapplicable in this problem. Why did he say we will be using it? this kind of error is really confusing; and, frankly, it's unacceptable that no correction has been made.
• There is no error. Sal uses the chain rule to compute the derivative of x with respect to t and the derivative of y with respect to t.

d/dt[x]=dx/dx•dx/dt by chain rule. But dx/dx=1, which is why he wrote in the multiplication by 1 in his calculation.
• Is there anywhere on Khan U where Sal provides a summary of the various "rules" (i.e. the power rule, the chain rule, the product rule, etc.)? If not, I think it would be beneficial to all of us students to publish a video with this information.
(1 vote)
• so what would the position vector be defined as? Since velocity is derived from postition
(1 vote)
• Super helpful!! Just wanted to say thanks
(1 vote)
• it seems like the initial path function is not a function of t--it would not make sense
to have x(t)*y(t)=16. So this is really different than anything done in the courses
up to this point, but there is no explanation. It seems the t parameter is being
introduced when differentiating with respect to t.
(1 vote)

## Video transcript

- [Voiceover] A particle moves along the curve xy equals 16, so that the y coordinate is increasing, we underline this, the y coordinate is increasing at a constant rate of two units per minute. That means that the rate of change of y with respect to t is equal to two. What is the magnitude, in units per minute, of the particle's velocity vector when the particle is at the point four comma four, so when x is four, and y is four. So let's see what's going on. So, let's first just remind ourselves what a velocity vector, what the velocity vector will look like. So our velocity is going to be a function of time. And it's going to have two components, it's going to be what is the rate of change in the x direction, and the rate of change in the y direction. So the rate of change in the x direction is going to be dx dt, and the rate of change in the y direction is going to be dy dt. And they tell us that this is, that dy dt is a constant two units per minute. But they're not even just asking us for just the velocity vector for its components, they're asking for the magnitude, they're asking for the magnitude of the particle's velocity vector. Well, if I have some vector, let me do a little bit of a side here, if I have some vector, let's say a, that has components b and c, well then the magnitude of my vector, sometimes you'll see it written like that, sometimes you'll see it written with double bars like that, the magnitude of my vector, and this comes straight out of the Pythagorean theorem, this is going to be the square root of b squared plus c squared. The square root of the x component squared plus the y component squared. So, if we wanted the magnitude of our velocity vector, the magnitude of the particle's velocity vector, well I could write that as the magnitude of v, I could even write it as a function of t, it's going to be equal to the square root of the x component squared, so that's the rate of change of x with respect to time, squared, plus the y component squared, which in this case is the rate of change of y, with respect to t, squared. So how do we figure out these, how do we figure out these two things? Well we already know the rate of change of y with respect to t, they say that's a constant rate of two units per minute. So we already know that this is gonna be two, or that this whole thing right over here's gonna be four, but how do we figure out the rate of change of x with respect to t? Well, we could take our original equation that describes the curve, we could take the derivative of both sides with respect to t, and then that's going to give us an equation that involves xy and dx dt and dy dt, so let's do that. So we have xy is equal to 16, I'm gonna to take the derivative with respect to t, of both sides, we do that in a different color, just for a little bit of variety, so the derivative with respect to t of the left hand side, derivative with respect to t of the right hand side, now the left hand side, we view this as a product of two functions, if we say, look, x is a function of t, and y is also a function of t, this is we're gonna do a little bit of the product rule and a little bit of the chain rule here. And so this is going to be equal to, derivative of the first function, which is, so, we'll first say the derivative of x with respect to x is one, times the derivative of x with respect to t. Remember, taking the derivative with respect to t, not with respect to x, times the second function, so times, times the second function, so times y, times y, plus, the first function, which is just x, times the derivative of the second function with respect to t. So first what's the derivative of y with respect to y, well that's just one, and then what's the derivative of y with respect to t, well that's dy dt, and that is going to be equal to, that is going to be equal to, derivative of a constant is just zero. So let's see, what does this simplify to, this simplifies to, in fact we don't even have to simplify it more, we can actually plug in the values to solve for dx dt. We know that dy dt is a constant two, and we want the magnitude of the particle's velocity vector when the particle's at the point four comma four, so when x is equal to four, so when x is equal to four, and y is equal to four, and y is equal to four. So now, it's a little messy right now, but this right here is an equation we can solve for, there's only one unknown here, the rate of change of x with respect to t, right when we are at the point four comma four, and so if we're able to figure that out, we can substitute that in here and figure out the magnitude of our velocity vector. So let us write it out, so this gives us four, four dx dt, plus what is this four times two, plus eight is equal to zero, and so we have four dx dt is equal to negative eight, just subtracted eight from both sides, divide both sides by four, you get dx dt, scroll down, is equal to negative two. So when all this stuff is going on the rate of change of x with respect to t is negative two, and then you square it, you get a four, right over here, and so the magnitude of our velocity vector is going to be equal to the square root of four plus four, which is equal to eight, which is the same thing as four times two, so this is going to be two square root of two, units per minute, so that's the magnitude of the velocity vector.