## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - Chapter Review Exercises - Page 638: 13

#### Answer

$$\frac{d y}{d x}= \frac{2t}{3t^2+1}$$ and at $t=3$, we have $$\frac{d y}{d x}= \frac{3}{14}.$$

#### Work Step by Step

Since $x=t^3+t$ and $y=t^2-1$, then we have $$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{y^{\prime}(t)}{x^{\prime}(t)}=\frac{2t}{3t^2+1}$$ and at $t=3$, we have $$\frac{d y}{d x}= \frac{6}{28}=\frac{3}{14}.$$

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