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Studying for a test? Prepare with these 6 lessons on Applications of derivatives.
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# Motion along a curve: finding rate of change

Video transskription
- [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, sub-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 as 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. Sub-units of distance per minute. And we are done.