Question

The following two sets of parametric functions both represent the same ellipse, Explain the difference between the graphs.x = 3 cost and y = 8 sin tx= 3 cos 4t and y = 8 sin 4t

Answer 1

Answer:

The graphs of both equations will have different slopes and periods

Explanation:

Given:

$\begin{gathered} x=3\cos t\text{ and }y=8\sin t \\ x=3\cos4t\text{ and }y=8\sin4t \end{gathered}$

Recall that the equation of an eclipse is generally given as;

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

We can go ahead and express both equations in the above as seen below;

$\begin{gathered} x^2=3^2\cos^2t\text{ and }y^2=8^2\sin^2t \\ \frac{x^2}{3^2}+\frac{y^2}{8^2}=\cos^2t+\sin^2t \\ \frac{x^2}{3^2}+\frac{y^2}{8^2}=1 \\ OR \\ x^2=3^2\cos^24t\text{ and }y^2=8^2\sin^24t \\ \frac{x^{2}}{3^{2}}+\frac{y^{2}}{8^{2}}=\cos^24t+\sin^24t \\ \frac{x^2}{3^2}+\frac{y^2}{8^2}=1 \end{gathered}$

So we can see that both equations represent the same eclipse.

Recall the below sine function;

$y=a\sin(bx-c)+d$

where;

a = amplitude

2pi/b = period

If we compare the given equations, we can see that x = 3 cos t and y = 8 sin t, have 3 and 8 as amplitudes respectively and pi as period.

While x= 3 cos 4t and y = 8 sin 4t have 3 and 8 as amplitudes respectively and pi/4 as period.

If we express the equations as linear functions we'll have;

For x = 3 cos t and y = 8 sin t;

$\begin{gathered} \frac{y}{x}=\frac{8\sin t}{3\cos t} \\ y=\frac{8}{3}x\tan t \end{gathered}$$\begin{gathered} \frac{y}{x}=\frac{8\sin t}{3\cos t} \\ y=\frac{8}{3}\tan t*x \end{gathered}$

For x = 3 cos 4t and y = 8 sin 4t

$\begin{gathered} \frac{y}{x}=\frac{8\sin4t}{3\cos4t} \\ y=\frac{8}{3}\tan4t*x \end{gathered}$

We can see that both equations have different slopes