Question

1. - / 0.81 points

find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.
$x = t\cos(t)$, $y = t\sin(t)$; $t = \pi$
$y = $

1. - / 0.81 points

find $dy/dx$ and $d^{2}y/dx^{2}$.
$x = e^{t}$, $y = te^{-t}$
$\frac{dy}{dx} = $
$\frac{d^{2}y}{dx^{2}} = $
for which values of $t$ is the curve concave upward? (enter your answer using interval notation.)

Explanation:

Problem 15

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

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:

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

Problem 16