Chapter 10 - Section 10.2 - Calculus with Parametric Curves - 10.2 Exercises - Page 655: 4

$y = 24x-40$

$x = \sqrt{t}$ $y = t^2-2t$ We can find $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t-2}{\frac{1}{2\sqrt{t}}} = 4\sqrt{t}~(t-1)$ When $t=4$: $x = \sqrt{4} = 2$ $y = (4)^2-2(4) = 8$ $\frac{dy}{dx} = 4\sqrt{4}~(4-1) = 24$ We can find the equation of the tangent line: $(y-8) = 24(x-2)$ $y = 24x-40$