W12 Regular

Due date: Sunday 4/5, 11:59pm

Parametric curves

01 $★$

02

Convert parametric curve to function graph

Write the following curves as the graphs of a function $y=f(x)$. (Find $f(x)$ for each case.)

(a) $x=t$, $y=t$ and $-\infty <t<\infty$

(b) $x=e^t$, $y=e^t$ and $-\infty <t<\infty$

Sketch each curve.

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Solution

Solutions - 0265-02

(a)

Observe that $x=t$ and $y=t$ implies $y=x$.

Therefore, all points on the curve satisfy $y=x$ and we set $f(x)=x$.

Since $x=t$ and $t$ covers the entire real line $x\in (-\infty ,\infty )$, the parametric curve is the entire line $y=x$.

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(b)

Observe that $x=e^t$ and $y=e^t$ implies $y=x$. Again, all points on the curve satisfy $y=x$ and so $f(x)=x$.

However, this time $t\in (-\infty ,\infty )$ implies $x>0$, and the entire range of $x>0$ is possible (set $t=\ln x$) to find an inverse.

So the image of this parametric curve is $y=x$ for $x>0$, and the origin is omitted.

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02 $★$

03

Convert function graph to parametric curve

Find parametric curves $c(t)=(x(t),y(t))$ whose images are the following graphs:

(a) $y=3x-4$ and $c(0)=(\mathrm{2,2})$

(b) $y=3x-4$ and $c(3)=(\mathrm{2,2})$

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Solution

Solutions - 0265-03

(a)

First choose a function $x(t)$, then set $y(t)=3x(t)-4$ to ensure the equation is satisfied.

When choosing $x(t)$, we want $x$ to cover the whole domain of $y=3x-4$ which is $x\in (-\infty ,\infty )$. We also need $x(0)=2$ to satisfy the initial condition.

Start by trying $x(t)=t$:

$$y=3t-4$$

But then $x(0)=-4$. Since $c(0)=(\mathrm{2,2})$ we should have $x(0)=2$. We can arrange for this by setting $x(t)=t+a$ and solving for $a$:

$$x(0)=2\gg \gg 0+a=2\gg \gg a=2$$

Therefore we define $x(t)=t+2$. Then:

$$y(t)=3(t+2)-4\gg \gg 3t+2$$

So we use:

$$\begin{array}{cc}x(t)& =t+2\\ y(t)& =3t+2\end{array}$$

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(b)

Same method but different condition:

$$x(3)=2\gg \gg 3+a=2\gg \gg a=-1$$

Therefore we define $x(t)=t-1$. Then:

$$y(t)=3(t-1)-4\gg \gg 3t-7$$

So we use:

$$\begin{array}{cc}x(t)& =t-1\\ y(t)& =3t-7\end{array}$$Link to original

Parametric calculus

03

03

Parametric concavity

Find the interval(s) of $t$ on which the parametric curve $c(t)=(t^2,t^3-4t)$ is concave up.

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Solution

Solutions - 0260-03

First derivative:

$$\begin{array}{c}\frac{dy}{dx}=\frac{y^{\prime }(t)}{x^{\prime }(t)}\\ \\ \gg \gg \frac{3t^2-4}{2t}\gg \gg \frac{3}{2}t-2t^{-1}\end{array}$$

Second derivative:

$$\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{1}{x^{\prime }(t)}\\ \\ \gg \gg \frac{d}{dt}(\frac{3}{2}t-2t^{-1})\cdot \frac{1}{2t}\\ \\ \gg \gg \frac{3}{4t}+\frac{1}{t^3}\gg \gg \frac{3t^2+4}{4t^3}\end{array}$$

This is positive if-and-only-if $t>0$. (Numerator always positive, denominator same sign as $t$.) So:

$$\text{Concave up:}t\in (0,+\infty )$$Link to original