W12 Stepwise

Due date: Friday 4/3, 11:59pm

Parametric curves

01

01

Convert parametric curve to function graph

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

(a) $x=t+3$, $y=4t$ and $0<t<1$

(b) $x=\cos t$, $y={\sin }^{2}t$ and $0<t<2\pi$

Sketch each curve.

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Solution

Solutions - 0265-01

(a) From the first equation, $t=x-3$.

Plug that in for $t$ in $y(t)$:

$$\begin{array}{c}y=4t\gg \gg y=4(x-3)\\ \\ \gg \gg y=4x-12\end{array}$$

The sketched curve should be the portion of the line with $x\in (\mathrm{3,4})$.

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(b)

For this one, best not to solve for $t$. Instead, notice the trig identity:

$$x^2+y={(\cos t)}^2+{\sin }^{2}t=1$$

Therefore the points on the curve satisfy the equation $x^2+y=1$. Solve this for the function:

$$y=1-x^2$$

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Parametric calculus

02

01

Parametric curves: Points with given slope

Where on the image of $(3t^2-2t,t^3-6t)$ does the tangent line have slope $3$?

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Solution

Solutions - 0260-01

Find a formula for the slope of the tangent line:

$$m=\frac{y^{\prime }(t)}{x^{\prime }(t)}\gg \gg \frac{3t^2-6}{6t-2}$$

Solve for the $t$ where $m=3$:

$$\begin{array}{c}\frac{3t^2-6}{6t-2}=3\gg \gg t^2-2=6t-2\\ \\ \gg \gg t^2-6t=0\\ \\ \gg \gg t=\mathrm{0,6}\gg \gg (x,y)=(\mathrm{0,0}),(\mathrm{96,180})\end{array}$$Link to original

03

02

Parametric concavity

Find ${\displaystyle \frac{d^{2}y}{d{x}^2}}$ at $t=1$ for the curve given parametrically by $x=4-t^{-2}$, $y=t^{-1}+t$.

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Solution

Solutions - 0260-02

Derivative functions:

$$(x^{\prime },y^{\prime })\gg \gg (2t^{-3},1-t^{-2})$$

Slope:

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

Second derivative:

$$\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{1}{x^{\prime }(t)}\\ \\ \gg \gg \frac{1}{2}(3t^2-1)\cdot \frac{1}{2t^{-3}}\gg \gg \frac{1}{4}(3t^5-t^3)\end{array}$$

At $t=1$:

$${\frac{d^{2}y}{d{x}^2}|}_{t=1}\gg \gg \frac{1}{4}(3-1)\gg \gg \frac{1}{2}$$Link to original