Welcome to a video on the Slope of Tangent Lines to Polar Curves.
The goals of this video are to determine dy/dx when an equation is given in polar form.
And then also to determine the slope of the tangent line to a polar curve.
So if r is equal to f of theta, or some function in terms of theta,
it probably makes sense from our study of parametric equations that dy/dx would equal dy/d theta divided by dx/d theta.
But to understand where this formula here comes from, we have to back to what we know about trigonometry.
Remember that x equals r cosine theta, and y is equal to r sin theta.
But since r is equal to f of theta, we can take r here and replace it with f of theta.
So we have x equals f of theta cosine theta.
We can replace this r with f of theta as well, so we have y equals f of theta sine theta.
So to determine dy/d theta,
we have to apply the product rule to f of theta sine theta,
which results in this sum here.
And to find dx/d theta, we have to apply the product rule to f of theta cosine theta.
Which would result in the difference of these two products because remember,
the derivative of cosine theta is equal to negative sine theta.
So you may want take a moment and see if you can derive this by applying the product rule.
But we’re gonna go ahead and take a look at a couple of examples.
So here we want to determine the slope of the tangent line to the plane curve at the following 3 points,
given that r equals 4.
So the first thing we have to realize is that if r is equal to 4, then f of theta is equal to 4.
So now we can go to our formula for dy/dx.
So dy/dx equals f prime of theta,
the derivative of 4 would be 0, so we’ll have 0 times sine theta plus f of theta, which is 4, times cosine theta
Divided by f prime of theta times cosine theta.
But again, f prime of theta would be 0.
So we have 0, cosine theta, minus f of theta, which is 4 sine theta.
So this would be 0, this would 0.
These fours simplify out, so we have cosine theta divided by negative sine theta.
So this would be negative cotangent theta.
So we have dy/dx equals negative cotangent theta.
So to determine the slope of the tangent line at these 3 points, we need to replace theta with the given angles.
So for this first point we’ll replace theta with pi over 4.
The second point we’ll replace theta with pi over 3.
And for the third point replace theta with pi over 2.
Remember that cotangent is the reciprocal of tangent.
And because the tangent of pi over 4 is equal to 1, so is the cotangent of pi over 4.
So here we have negative 1.
And then for the negative cotangent pi over 3, let’s go ahead and sketch a reference triangle.
It’s a 60-degree reference angle in the first quadrant.
So the cotangent of this angle would be the ratio of the adjacent side to the opposite side, 1 over the square of root 3.
So here we’ll have negative 1 over square root of 3, which simplifies to negative square root of 3 over 3.
And negative cotangent pi over 2 is equal to 0.
So these are the slopes of the tangent line at these 3 points.
So at (4, pi over 4) we’d be right here, so the slope of this tangent line should be negative 1.
Which looks like it would be.
At (4,pi over 3) we’d be somewhere in here, so our tangent line would look like this.
And negative square root of 3 over 3 is approximately equal to negative .5773.
So that looks about right.
And then lastly, at (4, pi over 2) we’d be right here.
And it’s pretty clear that this tangent line here would have a slope of 0.
Let’s go ahead and take a look at another example.
Here we want to determine the slope of the tangent line to the plane curve at the point (1,pi) given r equals 1 plus 3 sine theta.
Well the point (1,pi) would be right here in red, and the tangent line would look something like this.
And our goal is to find the slope of that tangent line by finding this derivative, and then replacing theta with pi.
Let’s go ahead and see if we can do that.
Remember, if r equals 1 plus 3 sine theta, that’s our f of theta.
So we’ll start with f prime of theta times sine theta.
Well f prime of theta is going to be 3 cosine theta. So we’ll have 3 cosine theta times sine theta.
Plus, f of theta times cosine theta.
Our denominator is going to be f prime of theta cosine theta.
So we’ll have 3 cosine theta times cosine theta minus f of theta times sine theta.
Now at this point we could just replace theta with pi and then evaluate this.
Or we could take the time to distribute here, combine like terms, and try to simplify the derivative.
Just for illustration purposes, let’s go ahead and see if we can simplify this derivative.
So we’ll start by distributing the cosine theta here and we’ll distribute the sine theta here.
So we’ll have plus cosine theta plus 3 cosine theta sine theta.
All over this would be 3 cosine squared theta.
Need to be careful here because of the subtraction.
We’re gonna have 1 times sine theta, sine theta.But it’s gonna be minus sine theta.
And then we’ll have 3 sine squared theta but will be minus 3 sine squared theta.
So not a lot simplifies here. We do have like-terms here in the numerator.
So we have dy/dx equals 6 cosine theta sine theta plus cosine theta.
All divided by 3 cosine squared theta minus sine theta minus 3 sine squared theta.
And we could perform some more trig substitution here, but let’s go and leave it like this.
And now replace theta with pi on the next screen.
So now we’ll replace theta with pi.
Cosine pi is going to be negative 1.
And the sine of pi is equal to 0.
And again, cosine pi would be negative 1.
In our denominator we’re gonna have 3 cosine squared pi, but cosine pi is equal to negative 1.
We’ll negative 1 squared minus sine pi would be 0.
Minus 3 sine squared pi, well sine pi is 0, so we have 0 squared.
So this simplifies to negative 1 all over positive 3.
So the slope of that tangent line is equal to negative one third on this point here.
And if we sketch this tangent line, a slope of negative one third looks accurate.
Okay that’s gonna do it for this video.
In the next video we’ll a take look at how to determine horizontal and vertical tangent lines.
I hope you found this helpful.