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?
I need to find an x and a y value, please help!
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