Answer
(a) $y=-x+\frac{5}{4}$
(b) $y=-x+\frac{5}{4}$
Work Step by Step
Part (a)
Find the slope of the tangent to the curve:
$m=\frac{dy}{dx}$
$m=\frac{dy/dt}{dx/dt}$
$m=\frac{-2\cos t\sin t}{\cos t}$
$m=-2\sin t$
$m=-2x|_{x=\frac{1}{2}}$
$m=-2\cdot \frac{1}{2}$
$m=-1$
Find the equation of the tangent to the curve at $(1/2,3/4)$:
$y-\frac{3}{4}=m(x-\frac{1}{2}$
$y-\frac{3}{4}=-1(x-\frac{1}{2})$
$y-\frac{3}{4}=-x+\frac{1}{2}$
$y=-x+\frac{5}{4}$
Part (b)
Eliminate the parameter:
$x=\sin t$ and $y=\cos^2t$
$x^2=\sin^2t$ and $y=\cos^2t$
$x^2+y=\sin^2t+\cos^2 t$
$x^2+y=1$
$y=-x^2+1$
Find the slope of the tangent to the curve at $(1/2,3/4)$:
$m=\frac{dy}{dx}$
$m=\frac{d}{dx}(-x^2+1)$
$m=-2x|_{x=\frac{1}{2}}$
$m=-2\cdot \frac{1}{2}$
$m=-1$
Find the equation of the tangent to the curve at $(1/2,3/4)$:
$y-\frac{3}{4}=m(x-\frac{1}{2}$
$y-\frac{3}{4}=-1(x-\frac{1}{2})$
$y-\frac{3}{4}=-x+\frac{1}{2}$
$y=-x+\frac{5}{4}$
