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Chapter 8: Problem 35

Determine the \(t\) intervals on which the curve is concave downward or concave upward. $$ x=t^{2}, \quad y=t^{3}-t $$

Short Answer

Expert verified
The curve is concave upward on \( (-\infty, 0) \cup (0,\infty) \). There is no region where the curve is concave downward.

Step by step solution

01

Find \(y'\)

We know that \(x = t^2\) and \(y = t^3 - t\). So, we can find \(y'\) with respect to \(t\): \(dy/dt = 3t^2 -1\), and \(dx/dt = 2t\). Therefore, \(y' = dy/dt/dx/dt = (3t^2 -1)/(2t)\).
02

Find \(y''\)

To find \(y''\), we need to take the derivative of \(y'\) using quotient rule: \((y'') = (d/dt(y')/(dx/dt) = ((6t^2)(2t)-(2(3t^2-1)))/(2t)^2 = 2t^2/(2t)^2\).
03

Determine intervals of concavity

Concavity is determined by the sign of the second derivative. If \(y'' > 0\), the curve is concave upwards. If \(y'' < 0\), the curve is concave downwards. Looking at \(y'' = 2t^2/(2t)^2\), we see that \(y'' > 0\) when \( t != 0\), and \(y''\) is undefined when \( t = 0\). Therefore, the curve is concave upward on \( (-\infty, 0) \cup (0,\infty) \) and there's no region where the curve is concave downward.

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Most popular questions from this chapter

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

In Exercises 43-46, find the area of the surface formed by revolving the curve about the given line. $$ \begin{array}{lll} \underline{\text { Polar Equation }} & \underline{\text { Interval }} & \underline{\text { Axis of Revolution }} \\ r=e^{a \theta} & 0 \leq \theta \leq \frac{\pi}{2} & \theta=\frac{\pi}{2} \end{array} $$

Explain why finding points of intersection of polar graphs may require further analysis beyond solving two equations simultaneously

In Exercises 43-46, find the area of the surface formed by revolving the curve about the given line. $$ \begin{array}{lll} \underline{\text { Polar Equation }} & \underline{\text { Interval }} & \underline{\text { Axis of Revolution }} \\ r=6 \cos \theta & 0 \leq \theta \leq \frac{\pi}{2} & \text { Polar axis } \end{array} $$

In Exercises 43-46, find the area of the surface formed by revolving the curve about the given line. $$ \begin{array}{lll} \underline{\text { Polar Equation }} & \underline{\text { Interval }} & \underline{\text { Axis of Revolution }} \\ r=a(1+\cos \theta) & 0 \leq \theta \leq \pi & \text { Polar axis } \end{array} $$
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