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The previous section discussed a special class of parametric functions called polar
functions. We know that
and so we can compute the derivative of with respect to using differentials: provided that .
With polar functions we have
provided that .
Consider the limaçon on the interval :
Find the equation of the tangent line to the curve at .
We start by
computing the derivative in rectangular coordinates. Recall that
and therefore When ,
Thus the equation of the line (in polarrectangular coordinates) tangent to the limaçon at is
Consider the limaçon given by on the interval . Find Cartesian coordinates
for the points where the tangent line to this limaçon is horizontal.
these points where the tangent lines is horizontal, we seek points where .
From earlier, we discovered so now we must find where the numerator is .
Since the numerator is equal to and since a product is zero when botheither of the terms is zero, the numerator is zero when
Let’s examine the case of first, and restrict to the situation where is between
and . If is in the interval , then when . If is in the interval , then when
The corresponding points in rectangular coordinates are
meaning , and
But the numerator was the product of and another term, namely So the numerator is also zero
At this point we have a choice to make: We can either apply the arcsin function and
solve for , or we can work with directly. Perhaps the easier route is to work
with directly. To explore , we may choose an algebraic method employing
the Pythagorean identity, or a geometric method looking at the unit circle
with the Pythagorean theorem. Let’s take the geometric route to compute .
Hence the derivative is zerois oneis undefined at
meaning , and
In summary, the graph of on the interval has horizontal tangent lines at the points
You can confirm this by looking at the graph below:
Again consider our friend the limaçon given by on the interval . Find Cartesian
coordinates for the points where the tangent line to this limaçon is vertical.
the vertical tangent lines, we seek points where the denominator of is equal to zero.
To start, we will convert to , and express the denominator as a quadratic expression
Since this is a quadratic equation in the variable , we invoke the quadratic formula to
Now we can find cosine using either geometrically via the unit circle or algebraically
with the Pythagorean identity. This time, let’s take the more algebraic route with the
Pythagorean identity. When , then
So and when ,
So So now we know that the graph of on has vertical tangent lines at the four
we see that the graph of on has vertical tangent lines at
You can confirm this visually by looking at the graph below:
When the graph of the polar function intersects the origin (sometimes called the
“pole”), then for some angle .
When , what is the formula for ?
The answer to the question above leads us to an interesting point. It tells us
the slope of the tangent line at the pole. When a polar graph touches the
pole at , the equation of the tangent line in polar coordinates at the pole is
Find the equations of the lines tangent to the graph of at the pole.
We need to know
Thus the equations of the tangent lines, in polarrectangular coordinates, are
In rectangular form, the tangent lines are and . You can confirm this by looking at
the graph below: