15. - / 0.81 points\nfind an equation of the tangent to the curve at the point corresponding to the given…

Problem 15

Explanation:

Step1: Find the point ((x,y))

When (t = \pi), (x=\pi\cos(\pi)=-\pi), (y = \pi\sin(\pi)=0).

Step2: Find (\frac{dy}{dx})

Use the formula (\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}). (\frac{dx}{dt}=\cos(t)-t\sin(t)), (\frac{dy}{dt}=\sin(t)+t\cos(t)). When (t = \pi), (\frac{dx}{dt}=\cos(\pi)-\pi\sin(\pi)=- 1), (\frac{dy}{dt}=\sin(\pi)+\pi\cos(\pi)=-\pi). So (\frac{dy}{dx}=\frac{-\pi}{-1}=\pi).

Step3: Use the point - slope form (y - y_0=m(x - x_0))

Here (m = \pi), (x_0=-\pi), (y_0 = 0). (y-0=\pi(x+\pi)), (y=\pi x+\pi^{2}).

Answer:

(y=\pi x+\pi^{2})

Problem 16

Explanation:

Step1: Find (\frac{dy}{dx})

Use the formula (\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}). (\frac{dx}{dt}=e^{t}), (\frac{dy}{dt}=e^{-t}-te^{-t}=(1 - t)e^{-t}). So (\frac{dy}{dx}=\frac{(1 - t)e^{-t}}{e^{t}}=(1 - t)e^{-2t}).

Step2: Find (\frac{d^{2}y}{dx^{2}})

Use the formula (\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}). (\frac{d}{dt}((1 - t)e^{-2t})=-e^{-2t}-2(1 - t)e^{-2t}=(2t - 3)e^{-2t}). So (\frac{d^{2}y}{dx^{2}}=\frac{(2t - 3)e^{-2t}}{e^{t}}=(2t - 3)e^{-3t}).

Step3: Find when the curve is concave upward

The curve is concave upward when (\frac{d^{2}y}{dx^{2}}>0). ((2t - 3)e^{-3t}>0). Since (e^{-3t}>0) for all (t), we solve (2t-3>0), (t>\frac{3}{2}).

Answer:

(\frac{dy}{dx}=(1 - t)e^{-2t}) (\frac{d^{2}y}{dx^{2}}=(2t - 3)e^{-3t}) ((\frac{3}{2},\infty))