(a) at what point does the curve

x = 1 - 2cos²(t), y = tan(t)(1 - 2cos²(t))
cross itself?

(x, y) =

(b) at the point where the above curve crosses itself, find the equations of both tangent lines. list them in order of in

tangent line with smaller slope: y = 2sin(2t)

tangent line with larger slope: y =

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Question

(a) at what point does the curve

x = 1 - 2cos²(t), y = tan(t)(1 - 2cos²(t))
cross itself?

(x, y) =

(b) at the point where the above curve crosses itself, find the equations of both tangent lines. list them in order of in

tangent line with smaller slope: y = 2sin(2t)

tangent line with larger slope: y =

note: you can earn partial credit on this problem.

preview my answers submit answers

your score was recorded.

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you have attempted this problem 2 times.

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Accepted Answer

### Part (a)
# Answer:
$(-1, 0)$
# Explanation:
## Step 1: Simplify $x$ and $y$
We know that $1 - 2\cos^{2}(t)=-\cos(2t)$ (using the double - angle formula $\cos(2t) = 2\cos^{2}(t)-1$, so $1 - 2\cos^{2}(t)=-(2\cos^{2}(t)-1)=-\cos(2t)$). So $x =-\cos(2t)$ and $y=\tan(t)\times(-\cos(2t))$

## Step 2: Find when the curve crosses itself
A curve given by parametric equations $x = x(t)$, $y = y(t)$ crosses itself when there exist two different values of $t$, say $t_1$ and $t_2$ ($t_1
eq t_2$) such that $x(t_1)=x(t_2)$ and $y(t_1)=y(t_2)$

We know that $\cos(2t)$ has a period of $\pi$, and $\tan(t)$ has a period of $\pi$

Let's consider $t$ and $t + \pi$ (since adding $\pi$ to the argument of $\tan$ gives the same value of $\tan$ and $\cos(2(t+\pi))=\cos(2t + 2\pi)=\cos(2t)$)

For $x$:
$x(t)=1 - 2\cos^{2}(t)$, $x(t + \pi)=1-2\cos^{2}(t+\pi)$
Since $\cos(t+\pi)=-\cos(t)$, then $\cos^{2}(t + \pi)=\cos^{2}(t)$, so $x(t)=x(t+\pi)$

For $y$:
$y(t)=\tan(t)(1 - 2\cos^{2}(t))$, $y(t+\pi)=\tan(t + \pi)(1-2\cos^{2}(t+\pi))$
Since $\tan(t+\pi)=\tan(t)$ and $\cos^{2}(t+\pi)=\cos^{2}(t)$, then $y(t)=y(t+\pi)$

Let's find the value of $x$ and $y$ when $t=\frac{\pi}{2}$ and $t = \frac{3\pi}{2}$ (or other suitable values). When $t=\frac{\pi}{2}$, $\cos(t) = 0$, $x=1-2\times0=- 1$, $\tan(\frac{\pi}{2})$ is undefined. Let's use the double - angle formula in another way.

We know that $1-2\cos^{2}(t)=-\cos(2t)$, so $x =-\cos(2t)$. When $2t=\pi$ (i.e., $t = \frac{\pi}{2}$), $x=-\cos(\pi)=1$? Wait, no. Wait, $1-2\cos^{2}(t)=-\cos(2t)$, when $2t = \pi$, $\cos(2t)=-1$, so $x=-(-1) = 1$? No, we made a mistake earlier. Let's use the identity $1-2\cos^{2}(t)=-\cos(2t)$ correctly. If $t=\frac{\pi}{4}$, $\cos(t)=\frac{\sqrt{2}}{2}$, $x = 1-2\times(\frac{\sqrt{2}}{2})^{2}=1 - 2\times\frac{1}{2}=0$, $y=\tan(\frac{\pi}{4})\times0 = 0$

Wait, let's start over. We know that $x = 1-2\cos^{2}(t)=-\cos(2t)$ (double - angle formula: $\cos(2t)=2\cos^{2}(t)-1\Rightarrow1 - 2\cos^{2}(t)=-\cos(2t)$)

We want to find $t_1$ and $t_2$ such that $x(t_1)=x(t_2)$ and $y(t_1)=y(t_2)$ with $t_1
eq t_2$

Let's consider the equation for $x$: $x = 1-2\cos^{2}(t)$, the range of $\cos^{2}(t)$ is $[0,1]$, so the range of $x$ is $[1 - 2\times1,1-2\times0]=[-1,1]$

We know that $y=\tan(t)x$ (since $y=\tan(t)(1 - 2\cos^{2}(t))=\tan(t)x$)

For the curve to cross itself, $y = 0$ (because if $y
eq0$, then $\tan(t)$ would have to be the same for two different $x$ values in a non - trivial way, but when $y = 0$, either $\tan(t)=0$ or $x = 0$)

If $y = 0$, then either $\tan(t)=0$ (i.e., $t = n\pi$, $n\in\mathbb{Z}$) or $x = 0$

If $t = n\pi$, $\tan(t)=0$, and $x=1-2\cos^{2}(n\pi)$

When $n$ is even, $n = 2k$, $\cos(2k\pi)=1$, $x=1 - 2\times1=-1$

When $n$ is odd, $n=2k + 1$, $\cos((2k + 1)\pi)=-1$, $\cos^{2}((2k + 1)\pi)=1$, $x=1-2\times1=-1$

When $x=-1$, $1-2\cos^{2}(t)=-1\Rightarrow2\cos^{2}(t)=2\Rightarrow\cos^{2}(t)=1\Rightarrow\cos(t)=\pm1\Rightarrow t = n\pi$

And $y=\tan(t)(1 - 2\cos^{2}(t))=\tan(n\pi)\times(-1)=0$

So when $t = 0$ and $t=\pi$ (two different values of $t$), $x=-1$ and $y = 0$. So the curve crosses itself at $(-1,0)$

### Part (b)
First, we need to find the derivative $\frac{dy}{dx}$ for the parametric curve. For a parametric curve $x = x(t)$, $y = y(t)$, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

1. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$
   - $\frac{dx}{dt}=\frac{d}{dt}(1 - 2\cos^{2}(t))$
     Using the chain rule, if $u=\cos(t)$, then $\frac{d}{dt}(1 - 2u^{2})=-4u\times(-\sin(t)) = 4\cos(t)\sin(t)=2\sin(2t)$ (using the double - angle formula $\sin(2t)=2\sin(t)\cos(t)$)
   - $\frac{dy}{dt}=\frac{d}{dt}[\tan(t)(1 - 2\cos^{2}(t))]$
     Using the product rule $(uv)^\prime=u^\prime v+uv^\prime$, where $u = \tan(t)$, $v=1 - 2\cos^{2}(t)$
     $u^\prime=\sec^{2}(t)$, $v^\prime = 2\sin(2t)$ (from the derivative of $x$ with respect to $t$)
     $\frac{dy}{dt}=\sec^{2}(t)(1 - 2\cos^{2}(t))+\tan(t)\times2\sin(2t)$
     Since $1 - 2\cos^{2}(t)=-\cos(2t)$ and $\sec^{2}(t)=1+\tan^{2}(t)$
     But we can also use $y=\tan(t)x$ and $x = 1-2\cos^{2}(t)$
     Alternatively, since we know that the curve crosses itself at $(-1,0)$, we need to find the values of $\frac{dy}{dx}$ at the two different $t$ values that give $x=-1,y = 0$

When $x=-1$, $1-2\cos^{2}(t)=-1\Rightarrow\cos^{2}(t)=1\Rightarrow\cos(t)=\pm1$
     Case 1: $\cos(t)=1$, then $t = 2k\pi$, $\sin(t)=0$, $\tan(t)=0$
       $\frac{dx}{dt}=2\sin(2t)=2\times2\sin(t)\cos(t)=0$ (wait, no, $\frac{dx}{dt}=4\cos(t)\sin(t)$, when $\cos(t)=1,\sin(t)=0$, $\frac{dx}{dt}=0$, we have a problem with the derivative. Wait, we made a mistake in the derivative of $x$. Let's recalculate $\frac{dx}{dt}$:
       $x = 1-2\cos^{2}(t)$, $\frac{dx}{dt}=-4\cos(t)(-\sin(t)) = 4\cos(t)\sin(t)=2\sin(2t)$
       When $\cos(t)=1$, $t = 2k\pi$, $\sin(2t)=\sin(4k\pi)=0$
       Case 2: $\cos(t)=-1$, then $t=(2k + 1)\pi$, $\sin(t)=0$, $\tan(t)=0$
       Wait, this is not the right approach. Let's use the fact that $y=\tan(t)x$ and $x = 1-2\cos^{2}(t)$

We can rewrite the parametric equations as $x=-\cos(2t)$, $y=-\tan(t)\cos(2t)$

Let's take $t=\frac{\pi}{2}$ and $t=\frac{3\pi}{2}$ (even though $\tan(\frac{\pi}{2})$ and $\tan(\frac{3\pi}{2})$ are undefined, we can take the limit as $t\to\frac{\pi}{2}$ and $t\to\frac{3\pi}{2}$)

As $t\to\frac{\pi}{2}^{-}$, $\tan(t)\to+\infty$, $x = 1-2\cos^{2}(t)\to1-2\times0 = 1$ (no, earlier mistake). Wait, let's go back to the correct point of self - intersection.

The correct point of self - intersection:
       We know that $x = 1-2\cos^{2}(t)=-\cos(2t)$, $y=\tan(t)(1 - 2\cos^{2}(t))=-\tan(t)\cos(2t)$

Let's consider $t=\frac{\pi}{4}$ and $t=\frac{5\pi}{4}$

For $t=\frac{\pi}{4}$:
       $x=1-2\times(\frac{\sqrt{2}}{2})^{2}=1 - 1 = 0$, $y=\tan(\frac{\pi}{4})\times0 = 0$
       For $t=\frac{3\pi}{4}$:
       $x=1-2\times(-\frac{\sqrt{2}}{2})^{2}=1 - 1 = 0$, $y=\tan(\frac{3\pi}{4})\times0 = 0$

Wait, we were wrong earlier. Let's use the double - angle formula correctly. The curve $x = 1-2\cos^{2}(t)=-\cos(2t)$, $y=\tan(t)(1 - 2\cos^{2}(t))=-\tan(t)\cos(2t)$

The function $x =-\cos(2t)$ has a period of $\pi$, and $y=-\tan(t)\cos(2t)$

When does $x(t_1)=x(t_2)$ and $y(t_1)=y(t_2)$ with $t_1
eq t_2$?

Let $t_1=\alpha$ and $t_2=\alpha+\pi$

$x(t_1)=-\cos(2\alpha)$, $x(t_2)=-\cos(2(\alpha+\pi))=-\cos(2\alpha + 2\pi)=-\cos(2\alpha)$

$y(t_1)=-\tan(\alpha)\cos(2\alpha)$, $y(t_2)=-\tan(\alpha+\pi)\cos(2(\alpha+\pi))=-\tan(\alpha)\cos(2\alpha)$

So $x(t_1)=x(t_2)$ and $y(t_1)=y(t_2)$ for $t_1$ and $t_1+\pi$

Let's find $x$ and $y$ when $\cos(2\alpha)=0$ (so that $x = 0$)

$\cos(2\alpha)=0\Rightarrow2\alpha=\frac{\pi}{2}+k\pi\Rightarrow\alpha=\frac{\pi}{4}+\frac{k\pi}{2}$

When $\alpha=\frac{\pi}{4}$, $x = 0$, $y=-\tan(\frac{\pi}{4})\times0 = 0$

When $\alpha=\frac{3\pi}{4}$, $x = 0$, $y=-\tan(\frac{3\pi}{4})\times0 = 0$

Now, let[SSE Completed, Client Connection Error][SSE onError error]

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step 1: Simplify \(x\) and \(y\)

Step 2: Find when the curve crosses itself

Part (b)

Answer: