Question

Find \frac{dy}{dx} and \frac{d^2y}{dx^2}.x=e^t,y=te^{-t}. For which values of t is the curve concave upward?

Differential equations
ANSWERED
asked 2021-05-03
Find \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}.x=e^t,y=te^{-t}\). For which values of t is the curve concave upward?

Expert Answers (2)

2021-05-04
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Best answer
2021-09-29

Answer:

\(y'=\frac{1-\ln x}{x^2}\)

and

\(y''=\frac{2\ln(x)-3}{x^3}\)

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

Explanation:

To determine the intervals of concavity, we apply the second derivative test to the function \(y=y(x)\), i.e. determine \(y''=\frac{d^2y}{dx^2}\)

To find the first and second derivative \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\)

as follows

\(y=te^{-t}\)

\(=\frac{t}{e^t}\)

\(=\frac{\ln(x)}{x}\)

where \(x=e^t\Rightarrow\ln(x)=\ln(e^t)=t\ln(e)=t\)

Thus, the parametric equations \(x=e^t\) and \(y=te^{-t}\) given equation (1) are represented by the cartesian equation (1) are represented by the cartesian equation

\(y=\frac{\ln x}{x}\)

where \(x=e^t\Rightarrow\ln(x)=\ln(e^t)=t\ln(e)=t\)

Thus, the parametric equations \(x=e^t\) and \(y=te^{-t}\) given in equation are represented by the cartesian equation

\(y=\frac{\ln(x)}{x}\)

To find the first derivative to the cartesian function we apply the quotient rule of derivatives as follows

\(\frac{d}{dx}[\frac{u(x)}{v(x)}]=\frac{u'v-v'u}{[v(x)]^2}\)

Thus,

\(y'=\frac{dy}{dx}=\frac{[\ln(x)]'x-[x]'\ln(x)}{[x]^2}\)

\(=\frac{[\frac{1}{x}]x-[1]\ln(x)}{[x]^2}\)

\(=\frac{1-\ln(x)}{x^2}\)

Again, we apply the quotient rule of derivatives to the function \(y'=\frac{1-\ln(x)}{x^2}\) as follows

\(y''=\frac{d^2y}{dx^2}=\frac{[1-\ln(x)]'x^2-[x^2]'[1-\ln(x)]}{[x^2]^2}\)

\(=\frac{[0-\frac{1}{x}]x^2-[2x][1-\ln(x)]}{x^4}\)

\(=\frac{-x-2x[1-\ln(x)]}{x^4}=\frac{x(-1-2[1-\ln(x)])}{x^4}\)

\(=\frac{-1-2[1-\ln(x)]}{x^3}=\frac{-1-2+2\ln(x)}{x^3}\)

\(=\frac{2\ln(x)-3}{x^3}\)

The relation the concavity of the graph of the function to its derivative is defined as follows

The graph of the function \(f(x)\) is concave upward if \(f''(x)>0\) and concave downward if \(f''(x)<0\)

Thus, from concavity theorem, we obtain

The graph of the function \(y=f(x)\) is concave upward when

\(y''=\frac{2\ln(x)-3}{x^3}>0\)

Thus, the inequality implies that

\(2\ln(x)-3>0\) and \(\frac{1}{x^3}>0\)

\(\ln(x)>\frac{3}{2}\) and \(\sqrt[3]{\frac{1}{x^3}}>\sqrt[3]{0}\)

\(\ln(x)>\frac{3}{2}\) and \(\sqrt[3]{\frac{1}{x^3}}>\sqrt[3]{0}\)

Thus we get

\(e^{\ln(x)}>e^{\frac{3}{2}}\) and \(\frac{1}{x}>0\)

\(x>e^{\frac{3}{2}}\) and \(\frac{1}{x}>0\)

Again, using the following parametric equation

\(x=e^t\)

Thus, the inequality implies that

\(x>e^{\frac{3}{2}}\) and \(\frac{1}{x}>0\)

\(e^t>e^{\frac{3}{2}}\) and \(\frac{1}{e^t}>0\)

\(e^t>e^{\frac{3}{2}}\) and \(e^{-t}>0\)

Take "ln" for both sides of the previous inequality. Thus we get

\(\ln(e^t)>\ln(e^{\frac{3}{2}})\) and \(\ln(e^{-t})>\ln(0)\)

\(t>\frac{3}{2}\) and \(-t>\ln(0)\)

\(t>\frac{3}{2}\) and \(t>-\ln(0)\)

The inequality \(t>-\ln(0)\) is not permissible since the domain of the logarithmic function consists of all positive real numbers \((0,\infty)\) and thus the following natural logarithmic function \(\ln(0)\) is not defined.

therefore, the curve is concave upward in the interval \(t\in(\frac{3}{2},\infty)\)

 

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