# Thread: Tangents - Parametric Equations

1. ## Tangents - Parametric Equations

Suppose a curve is traced by the parametric equations:

x = 4sin(t)

y = 47 - 16cos^2(t) - 32sin(t)

At what point on this curve is the tangent line horizontal?

2. Originally Posted by Del
Suppose a curve is traced by the parametric equations:

x = 4sin(t)

y = 47 - 16cos^2(t) - 32sin(t)

At what point on this curve is the tangent line horizontal?

The slope of a curve y(x), when expressed parametrically, may be found by
$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

So
$\displaystyle \frac{dy}{dx} = \frac{4~cos(t)}{32~sin(t)~cos(t) - 32~cos(t)} = 0$

$\displaystyle = \frac{4}{32~sin(t) - 32} = 0$

Notice that there is no value of t such that the derivative is 0. Thus there is no solution.

-Dan

3. I see, but I am looking for the point on the graph where the tangent would be horizontal, the x-value will be 4, but what will the y-value be?

Edit: I found it to be 15, thanks!

4. ## errors

$\displaystyle \frac{dy}{dx}=\frac{32\sin(t)\cos(t)-32\cos(t)}{4\cos(t)}=0$

--Kevin C.

5. Originally Posted by TwistedOne151
$\displaystyle \frac{dy}{dx}=\frac{32\sin(t)\cos(t)-32\cos(t)}{4\cos(t)}=0$
However, I still say there is no such point where the tangent is horizontal. The derivative is discontinuous at $\displaystyle t = \pi / 2$ isn't it? (I always get screwed up when a function "ends" suddenly.)