# Slopes of Polar Curves

Okay, now that we have described what
a polar curve is, curves and polar coordinates where we had the radius
as a function of theta, what we want to do is we want to do calculus with
these polar curves. We want to figure out how do you
get the slope of the tangent line, that is dy/dx if somebody just
gives you r as a function of ϴ. And the key is to think of it
as a parameterized curve. See if we just let, we can call
our parameter t and we just let ϴ=t and the r is f (ϴ) which is
just f (t) and that tells us what x and y are because x is
r cos (ϴ). That’s f(t) cos(ϴ). And y is r sin (ϴ).
So that’s f(t) sin(ϴ). So now that we’ve got x and y
as functions of t, we can take the derivatives. The derivative of y is f(t) cos(t).
This is derivative of sin is cos. Plus f'(t) sin(t). Just by the product
rule and the derivative of x is -f(t) sin(t) + f'(t) cos(t). And then if you know dy/dt and
dx/dt, you take their ratio and you get dy/dx. Okay, it’s an ugly formula but it’s
a formula and it’s often easier to just express everything back
again in terms of r and ϴ because after all, f(t) is the same
thing as r and cos(t) is the same thing as cos (ϴ) because t is just ϴ.
In fact, we could have called our parameter ϴ to begin with. But we’re
used to calling our parameter t so I wrote everything in terms of t.
So you will always want r cos (ϴ) + r’ and by r prime I mean
the derivative of r with respect to ϴ, and we have this whole expression.
So let’s use this on a couple of examples. So there we are.
Put a box around it. We will use this in a couple
of examples and figure out the slopes of some curves.
The first example is we wanna look at a circle. Just take a circle
of radius 2 and we wanna compute the slope of the tangent line
as a function of where you are on the circle. That’s pretty easy. r is 2 so
r’ is 0. Our formula says r cos (ϴ) + r’ sin (ϴ)
over that. r’ is 0. So you just get 2, r is canceled.
You just wind up with -cot (ϴ). This should make sense geometrically.
Because here we have our circle. And we are looking at what’s happening
at angle ϴ and the line that heads out like this, the slope of this line
is the tangent of ϴ. The line we are interested in
is perpendicular to that and it has a slope of – cot (ϴ).
You may remember that when you wanna take a perpendicular to a line,
you take minus the reciprocal of the slope, minus the reciprocal
of tangent of ϴ is – cot (ϴ). Okay, for our next example, we’re gonna
look at a curve that’s called a cardioid. A cardioid is r=1+sin(ϴ) and
it looks like this. If we run 1-sin(ϴ), it would have
looked like a heart, which is why it’s called a cardioid. Now it looks
like an upside down heart and we want to figure out what is the slope
of the tangent line when ϴ=π/4. Again, it’s just plug and chug. r is 1+sin(ϴ). The derivative of that is
r’=cos(ϴ) and then we just plug and chug. You have dy/dx. Here’s our formula. r is (1+sinϴ) times cosϴ,
r’ is cosϴ(sinϴ). We collect terms. We get this whole
ugly expression and in the end, that works out to -1 + √2. It’s not an especially enlightening
calculation. It’s a formula. You plug into the formula but if you can
do this problem, you can do almost any problem.