 Suppose a curve is given in parametric form. We can do calculus on the functions defining the coordinates, and so the natural question to ask is, could we find the slope of the tangent line to a curve given in parametric form? Well, I hope so, otherwise this will be a very short video. A useful thing to remember from single variable calculus is that if you want to find the derivative of the inverse function, then as long as we actually have an inverse, the derivative of x with respect to y is 1 divided by the derivative of y with respect to x. And that's useful as long as we actually have the inverse function. So another useful thing to remember is that suppose the derivative exists and doesn't change signs in some interval between a and b, then our function is invertible in that interval. And since for most functions the sign changes only occur at discrete points, this means that in general we won't have to worry about whether the inverse function exists because it will exist almost everywhere. So how does this apply to curves defined parametrically? Well, suppose x is some function of t, and y is also some function of t. If I want to find dy dx, the chain rule tells me that that's going to be dy dt times dt dx. And again, x is probably invertible anywhere we care to look, so this dt dx, well, that's really the same as 1 divided by dx dt, or dy dt divided by dx dt. And this leads us to the result, let x be some function of t, let y be some function of t, then dy dt is the quotient of the derivatives. For example, suppose x of t and y of t is some curve where x of t is t squared and y of t is sin of t. And let's do a standard single variable calculus problem and find the equation of the line tangent to the graph at t equal to pi over 4. So the first thing we need to write the tangent line is a point on the curve, and at t equal pi over 4 our x and y values will be... Now if I want to write the equation of the tangent line I do need to find that derivative. So dy dx, well, that's the ratio, the quotient of dy dt and dx dt. So let's differentiate both x of t and y of t. And so dy dx will be... And at t equal to pi fourths dy dx will be... And so I have a point on the line, I have the derivative and I can now find the tangent line. And so given a curve to find parametrically we could find the slope of the line tangent to the graph at any point, which is extremely useful. But it turns out this is about the least interesting thing we can do when we have our curve to find parametrically because we can do so much more once we have those parametric equations. We'll take a look at that next.