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Current time:0:00Total duration:5:52

Motion along a curve: finding rate of change

AP.CALC:
FUN‑8 (EU)
,
FUN‑8.B (LO)
,
FUN‑8.B.1 (EK)

Video transcript

we're told that a particle moves along the curve x squared y squared is equal to 16 so that the x-coordinate is changing at a constant rate of of negative 2 units per minute what is the rate of change in units per minute of the particles y-coordinate when the particle is at the point 1 comma 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 let me underline that the x coordinate is changing at a constant rate of negative 2 units per minute so we could say that DX I'll write it 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 2 and they're saying units be some units some unit of distance units divided by minute units per minute and what they want us to figure out is what is the rate of change of the particles y-coordinate so let me underline that what is the rate of change of the particles y-coordinates so what they want us to find is what is dy DT what is that equal to and we and they say when the particle is at the point 1 comma 4 so when X is equal to 1 so X is equal to 1 and Y is equal to 4 y is equal to 4 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&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 gonna take the derivative actually let me just let me just you let me just erase this so I have a little bit more space all right and so that way I can just add it so let's take the derivative with respect to T of both sides of that and if 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 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 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 T and so once again we're going to apply the chain rule the derivative of Y squared with respect to Y is 2y let me 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 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 and so here we have it we need to simplify this a little bit but we have an equation that gives a 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 2x y squared DX DT a plus actually I don't even have to rewrite it again we can all we're doing 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 1 so we know that the X's here are equal to 1 this X x squared well that's just going to be 1 squared so that's going to be equal to 1 we know that Y is equal to four so this is going to be 16 and this is going to be 8 we know the derivative of X with respect to T is negative 2 they tell us that in the problem statement negative 2 and 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 2 times 1 times negative 2 so that is negative 4 times 16 so that is negative 64 and then we have and then we have let me do this in a color you can see and then we have all of this well this is just going to be 1 times 8 times dy DT so this is going to be 8 dy DT so plus 8 times the derivative of Y with respect to T is equal to 0 add 64 to both sides and we get I'll switch to a neutral color 8 times the derivative of Y with respect to T is equal to 64 divide both sides by 8 and you get the derivative of Y with respect to T is equal to 64 divided by 8 is just 8 and if you want if you want look at the units it will also be in units per minute use some units of distance per per minute and we are done