## Calculus 8th Edition

$y=24x-40$
$\qquad\qquad\qquad\qquad\qquad\qquad$Find the point at which $t = 4$ $x|_{t=4}=\sqrt{4}=2$ $y|_{t=4}=(4)^2-2(4)=16-8=8$ Point: $(2,8)$ $\qquad\qquad\qquad\qquad\qquad\qquad\qquad$Now find $\displaystyle \frac{dy}{dx}$ $\displaystyle \frac{dy}{dt}=2t-2$ $\displaystyle \frac{dx}{dt}=\frac{1}{2\sqrt{t}}$ $\displaystyle \frac{dy}{dx}=\frac{2t-2}{\frac{1}{2\sqrt{t}}}=2\sqrt{t}(2t-2)=4\sqrt{t}(t-1)$ $\qquad\qquad\qquad\qquad$Now plug $t=4$ in for $t$ and find the slope of the line: $\displaystyle \frac{dy}{dx}|_{t=4}=4\sqrt{4}(4-1)=4(2)(3)=24$ $\qquad\qquad\qquad\qquad$Now use the point-slope form of a line to $\qquad\qquad\qquad\qquad$find the equation tangent to the parametric curve. $y-y_1=m(x_1-x)\quad$ $y-8=24(x-2)$ $y=24x-48+8$ $y=24x-40$