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Chapter 8: Problem 38

Determine the \(t\) intervals on which the curve is concave downward or concave upward. $$ x=t^{2}, \quad y=\ln t $$

Short Answer

Expert verified
The curve is concave downward for all \(t > 0\).

Step by step solution

01

Find first derivative

Start with the parametric definitions \(x = t^2\) and \(y = \ln t\). To get dy/dx, we need to find dx/dt and dy/dt first. The derivative of \(x\) with respect to \(t\) is \(dx/dt = 2t\) while the derivative of \(y\) with respect to \(t\) is \(dy/dt = 1/t\). Divide \(dy/dt\) by \(dx/dt\) to get: \[dy/dx = \frac{1/t}{2t} = \frac{1}{2t^2}\].
02

Find second derivative

The second derivative \(d^2y/dx^2\) is given by differentiating dy/dx with respect to 't' and then dividing by \(dx/dt = 2t\). The derivative of \(\frac{1}{2t^2}\) with respect to 't' is \(-\frac{1}{t^3}\). Therefore, \[d^2y/dx^2 = \frac{-1/t^3}{2t} = -\frac{1}{2t^4}\].
03

Determine the intervals of concavity

The curve is concave upward wherever the second derivative \(d^2y/dx^2 = -1/2t^4 > 0\), and concave downward wherever the second derivative is less than zero. But here, the second derivative is always negative for all real positive t (excluding 0), thus the curve is concave downward for all \(t > 0\).

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Most popular questions from this chapter

In Exercises \(27-38,\) find a polar equation for the conic with its focus at the pole. (For convenience, the equation for the directrix is given in rectangular form.) \(\frac{\text { Conic }}{\text { Hyperbola }} \quad \frac{\text { Eccentricity }}{e=\frac{3}{2}} \quad \frac{\text { Directrix }}{x=-1}\)

In Exercises \(7-16,\) find the eccentricity and the distance from the pole to the directrix of the conic. Then sketch and identify the graph. Use a graphing utility to confirm your results. \(r=\frac{3}{2+6 \sin \theta}\)

Write a short paragraph describing how the graphs of curves represented by different sets of parametric equations can differ even though eliminating the parameter from each yields the same rectangular equation.

Show that the polar equation for \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is \(r^{2}=\frac{b^{2}}{1-e^{2} \cos ^{2} \theta} \cdot \quad\) Ellipse

Use a graphing utility to graph the polar equation over the given interval. Use the integration capabilities of the graphing utility to approximate the length of the curve accurate to two decimal places. $$ r=2 \sin (2 \cos \theta), \quad 0 \leq \theta \leq \pi $$
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