a) Parametrize the given ellipse (x/4)^{2} + (y/2)^{2} = 1 What will the parametrization be if...


a) Parametrize the given ellipse

$$\left( \frac{x}{4} \right)^{2} + \left( \frac{y}{2} \right)^{2} = 1 $$

What will the parametrization be if the center of the ellipse is translated to the point {eq}(5, 1) {/eq}?

b) Find parametric equations for the segment joining the points {eq}(2, 3) {/eq} and {eq}(4, 6), \, 0 \leq t \leq 1 {/eq}.

c) Use the formula for the slope of the tangent line to find {eq}\frac{dy}{dx} {/eq} for the curve {eq}c(t) = (7t^{3}, 4t^{2} - 1) {/eq} at {eq}t = 4 {/eq}.

Parametric Equations:

In this series of semi-related problems (that last one is a complete dark horse here), we will be dealing with parameterizations. In the first two, we will parameterize some curves. In the first case, we will apply a polar parameterization, and in the second we will use a trick that is super straightforward. We'll deal with that dark horse when we get to it.

Answer and Explanation:

Part A

We can use a polar parameterization where we simply have different radii along each axis. The axis along {eq}x {/eq} has radius 4 and the axis along {eq}y {/eq} has radius 2, so a parameterization is

{eq}\begin{align*} x &= 4 \cos t \\ y &= 2 \sin t \\ t &\in [0, 2\pi] \end{align*} {/eq}

If we move the center to (5,1), we just need to make sure we are always adding 5 to {eq}x {/eq} and 1 to {eq}y {/eq}, so the parameterization is simply

{eq}\begin{align*} x &= 4 \cos t + 5 \\ y &= 2 \sin t + 1 \\ t &\in [0, 2\pi] \end{align*} {/eq}

Part B

Parameterizing a line segment on {eq}t \in [0,1] {/eq} is one of the easiest parameterizations ever. All we have to do is think about what we need to do to the current point to get it to the next (one way to think about the parameter in this case is as an off/on switch). We need to add 2 to {eq}x {/eq} to get it from 2 to 4, and we need to add 3 to {eq}y {/eq} to get it from 3 to 6, so a parameterization on {eq}t \in [0,1] {/eq} is

{eq}\begin{align*} x &= 2+2t \\ y &= 3+3t \end{align*} {/eq}

Part C

Now this one. First, we do not need to memorize a formula to do this! The fact that we are being asked to use a formula here is unfortunate, as it seems to imply a certain level of difficulty required to obtain the formula, and robs us of any understanding of what is actually taking place. All we need to do here is apply the chain rule, which is really pretty obvious since we have both {eq}x {/eq} and {eq}y {/eq} in term of {eq}t {/eq}. How else are we going to change those variables? All we do is rewrite the derivative {eq}\frac{dy}{dx} {/eq} with it:

{eq}\begin{align*} \frac{dy}{dx} &= \frac{dy/dt}{dx/dt} \end{align*} {/eq}

This is the "formula" we want to use it. We get

{eq}\begin{align*} \frac{dy}{dx} &= \frac{dy/dt}{dx/dt} \\ &= \frac{\frac{d}{dt} \left( 4t^2-1 \right)}{\frac{d}{dt} \left( 7t^3 \right)} \\ &= \frac{8t}{21t^2} \\ &= \frac8{21t} \end{align*} {/eq}

So when {eq}t = 4 {/eq} the slope of the curve is

{eq}\begin{align*} y'(4) &= \frac8{21(4)} \\ &= \frac2{21} \\ &\approx 0.095 \end{align*} {/eq}

Learn more about this topic:

Evaluating Parametric Equations: Process & Examples

from Precalculus: High School

Chapter 24 / Lesson 3

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