Question

1. - / 1.66 points find \\( \frac{d y}{d x} \\) and \\( \frac{d^{2} y}{d x^{2}} \\). \\( x=e^{t}, \quad y=t e^{-t} \\) \\( \frac{d y}{d x}= \\) \\( \frac{d^{2} y}{d x^{2}}= \\) for which values of \\( t \\) is the curve concave upward? (enter your answer using interval notation.)

Explanation:

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

Using the derivative rules:

• For \(x = e^{t}\), \(\frac{dx}{dt}=e^{t}\) (since \(\frac{d}{dt}(e^{at}) = ae^{at}\), here \(a = 1\)).
• For \(y=te^{-t}\), use the product rule \((uv)^\prime=u^\prime v + uv^\prime\) where \(u = t\), \(u^\prime=1\), \(v = e^{-t}\), \(v^\prime=-e^{-t}\). Then \(\frac{dy}{dt}=e^{-t}-te^{-t}=(1 - t)e^{-t}\).

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

By the chain - rule \(\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
Substitute \(\frac{dy}{dt}=(1 - t)e^{-t}\) and \(\frac{dx}{dt}=e^{t}\) into the formula:
\(\frac{dy}{dx}=\frac{(1 - t)e^{-t}}{e^{t}}=(1 - t)e^{-2t}\).

Step3: Find \(\frac{d^{2}y}{dx^{2}}\)

First, find \(\frac{d}{dt}(\frac{dy}{dx})\).
Let \(u=(1 - t)\) and \(v = e^{-2t}\). Using the product rule \((uv)^\prime=u^\prime v+uv^\prime\), where \(u^\prime=-1\) and \(v^\prime=-2e^{-2t}\).
\(\frac{d}{dt}(\frac{dy}{dx})=-e^{-2t}-2(1 - t)e^{-2t}=(2t - 3)e^{-2t}\).
Then, by the formula \(\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\).
Substitute \(\frac{d}{dt}(\frac{dy}{dx})=(2t - 3)e^{-2t}\) and \(\frac{dx}{dt}=e^{t}\) into the formula:
\(\frac{d^{2}y}{dx^{2}}=(2t - 3)e^{-3t}\).

Step4: Determine when the curve is concave upward

The curve is concave upward when \(\frac{d^{2}y}{dx^{2}}>0\).
Since \(e^{-3t}>0\) for all \(t\in R\) (because the exponential function \(y = e^{ax}\), \(a=-3\) has \(y>0\) for all \(x\)), we solve the inequality \(2t-3>0\).
\(2t-3>0\Rightarrow2t>3\Rightarrow t>\frac{3}{2}\).

Answer:

\(\frac{dy}{dx}=(1 - t)e^{-2t}\)

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

The curve is concave upward for \(t\in(\frac{3}{2},\infty)\)