# Hyperbolic coordinates (R, \theta) of a given point \mathcal{P} in the xy plane by...

## Question:

Hyperbolic coordinates {eq}(R, \theta) {/eq} of a given point {eq}\mathcal{P} {/eq} in the {eq}xy {/eq} plane by requiring that the Cartesian description {eq}(x,y) {/eq}of {eq}\mathcal {P} {/eq} be given by the relationship:

{eq}\begin {cases} x = R \cosh \theta \\ y = R \sinh \theta \end {cases} {/eq}

A curve in hyperbolic coordinates is given by {eq}R(\theta) = \theta +1 {/eq}.

Find {eq}\frac {dy}{dx} {/eq} when {eq}\theta=0. {/eq}

## Derivatives of Parametric Equations

Given an equation parametrically defined as {eq}(x,y) = (f(a,b),\ g(a,b)) {/eq}, the derivative {eq}\frac{dy}{dx} {/eq} is equal to the ratio of the total partial derivative of {eq}y {/eq} with respect to {eq}a {/eq} and {eq}b {/eq}, and the total derivative of {eq}x {/eq} with respect to {eq}a {/eq} and {eq}b {/eq}.

## Answer and Explanation:

Given that {eq}R(\theta) = \theta + 1 {/eq}, then the point {eq}(x,y) {/eq} corresponds to the point {eq}((\theta+1)\cosh\theta,\ (\theta+1) \sinh\theta) {/eq}. Hence,

$$\begin{align*} \frac{dy}{dx} &= \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\\ &=\frac{\frac{d}{d\theta}\left ( (\theta + 1)\sinh\theta \right )}{\frac{d}{d\theta}((\theta + 1)\cosh\theta)}\\ \\ \frac{d}{d\theta}\left ( (\theta + 1)\sinh\theta \right ) &= \frac{d}{d\theta}(\theta+1)\sinh\theta + (\theta + 1)\frac{d}{d\theta}(\sinh\theta) \\ &= \sinh\theta + (\theta + 1)(\cosh\theta)\\ \\ \frac{d}{d\theta}\left ( (\theta + 1)\cosh\theta \right ) &= \frac{d}{d\theta}(\theta+1)\cosh\theta + (\theta + 1)\frac{d}{d\theta}(\cosh\theta) \\ &= \cosh\theta + (\theta + 1)(\sinh\theta)\\ \\ \frac{dy}{dx} &= \frac{\sinh\theta + (\theta + 1)(\cosh\theta)}{\cosh\theta + (\theta + 1)(\sinh\theta)} \end{align*} $$

At {eq}\theta = 0 {/eq}:

$$\begin{align*} \frac{dy}{dx} &= \frac{\sinh(0)+ (0 + 1)(\cosh(0))}{\cosh(0) + (0 + 1)(\sinh(0))}\\ &= \frac{0+1}{1+0}\\ &= \boxed{1} \end{align*} $$

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from Precalculus: High School

Chapter 24 / Lesson 3