 To find the tangent to a polar curve is difficult if we use polar coordinates. The problem is that the line through a point has a simple equation of rectangular coordinates, but as we saw, simple equations of rectangular coordinates usually become complicated equations in polar coordinates. However, a problem exists whether or not we can solve it easily, so let's consider the problem of finding the tangent to a curve defined in polar coordinates. So let's try to find the slope of the line tangent to the graph of r equals 2 minus cosine theta at theta equal pi fourths. We could try to convert this into a rectangular equation and compute the derivative normally. We have our conversion formulas, and so our equation becomes, which we differentiate to get a real mess. I really don't want to have to differentiate this, so maybe we don't want to do it this way. So let's try a parametric approach. So remember the point r theta in polar coordinates is the point x, y in rectangular, where x is our cosine theta and y is our sine theta. We can treat a curve defined using polar coordinates as a curve defined parametrically. x of theta is r of theta cosine theta and y of theta is r of theta sine theta. This allows us to find the derivatives of x and y with respect to theta, and we can find dy dx as the quotient, which will give us the slope of the tangent line. At this point, things get messy. So we have y equals r sine theta. So y of theta is r, that's 2 minus cosine theta times sine theta. We can differentiate, and at pi fourths we find, similarly, x is r cosine theta. Again, r is 2 minus cosine theta, so x of theta is, and the derivative will be, and at pi fourths we'll get, so we know both y prime and x prime, so dy dx will be the quotient, and we need to find the point on the graph, so we find that, at theta equals pi fourths, we're at, and so this tangent line will pass through the point, square root 2 minus 1 half, square root 2 minus 1 half, and so its equation will be. Now, stylistically, because the problem was given in polar coordinates, we should actually answer in polar coordinates as well. And so in polar coordinates, this equation will become, and we could solve this equation for r, but we have plans for this weekend, so we'll leave it as it is. Now we can find it the slope of the tangent line to the graph of r equals r of theta by using the formula, actually don't bother memorizing a formula, instead rely on this basic idea of treating both x and y as functions of theta, finding the x d theta and dy d theta, and then finding the derivative dy dx as the quotient of these two derivatives. The reason it's important not to try to find a formula is your formulas won't work for many graphs defined in polar coordinates, so for example, let's try to find the slope of the line tangent of the graph of r squared equals sine 2 theta at any point r theta. Now it turns out we'll want to find dr d theta. Since r squared equals sine 2 theta, we can find dr d theta using implicit differentiation, which will give us, again, y of theta, we'll assume that's r of theta sine of theta, so y prime will be, and we know what r prime of theta is, and so we can replace. Similarly, we'll assume that x of theta is r theta cosine theta, so we find the derivative will be, and we have y prime and x prime, so dy dx will be the quotient. And because this is a compound fraction, we should probably simplify this if we multiply numerator and denominator by 2r we'll get, and we can go one step further. We know that r squared is sine of 2 theta, and so we can get our final answer. And if you really want to impress your friends, you can probably simplify this even further by expanding sine 2 theta and cosine 2 theta and using a lot of trigonometric identities, but again, we have plans for the weekend.