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Current time:0:00Total duration:6:22

Video transcript

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 2 units per minute that means that the rate of change of Y with respect to T is equal to 2 what is the magnitude in units per minute of the particles velocity vector when the particle is at the point 4 comma 4 so when X is 4 and Y is 4 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 2 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 particles velocity vector well if I have some vector let me do a little bit of an aside here if I have some vector let's say a that is it has components I don't know B and C well then the magnitude of my vectors 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 want to know the magnitude of our velocity vector the magnitude of the particles velocity vector well I could write that as the magnitude of V and 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 square plus 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 going to be 2 or that this whole thing right over here is going to be 4 but how do we figure out the rate of change of X with respect to T well we could we could take our original 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 X Y and then DX DT and dy DT so let's do that so we have x y is equal to 16 I'm gonna take the derivative with respect to T of both sides let me do that in a different color just for a little bit of variety so a 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 the 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 a little bit of the chain rule here and so this is going to be equal to derivative of derivative of the first function which is so we'll first say the derivative of X with respect to X is 1 times the derivative of X with respect to T remember we're taking the derivative with respect to T not with respect to x times the second function so x 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 1 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 - and we want the magnitude of the particles velocity vector when the particle is at the point 4 comma 4 so when X is equal to 4 so when X is equal to 4 and Y is equal to 4 and Y is equal to 4 so now it's a little messy right now but this right here is an equation that 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 4 comma 4 and so if we're able to figure that out we could substitute that in here and figure out the hole the magnitude of our velocity vector so let us write it out so this gives us 4 for DX DT plus what is this 4 times 2 plus 8 is equal to 0 and so we have 4 DX DT is equal to negative 8 - subtracting 8 from both sides divide both sides by 4 you get DX DT let me scroll down is equal to negative 2 so when all the stuff is going on the rate of change of X with respect to T is negative 2 and so then you square it you get a 4 right over here and so the magnitude of our velocity vector is going to be equal to equal to the square root of 4 plus 4 which is equal to 8 which is the same thing as 4 times 2 so this is going to be 2 square roots of 2 units per minute so that's the magnitude of the velocity vector
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