8.1 Slopes and Tangents for the Parametric Curves
If $x$ and $y$ are both given as functions of a third variable $t$, then the equations
$
x=f(t), \quad y=g(t)
$
are called parametric equations, and $t$ is called the parameter.
The set of points $(x, y)=(f(t), g(t))$ defined by the parametric equations is called the parametric curve.
When the points in a parametric curve are plotted in order of increasing values of $t$, the curve is traced out in a specific direction. This is called the direction of path (or motion) of the curve.
Parametric Formula for $d y / d x$
If the equation $x=f(t), y=g(t)$ define $y$ as a differentiable function of $x$ and $d x / d t \neq 0$, then
$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$, where $\frac{d x}{d t} \neq 0$.
Parametric Formula for $\frac{d^2 y}{d x^2}$
$\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{\frac{d}{d t}\left(\frac{d y}{d x}\right)}{\frac{d x}{d t}}$
Horizontal Tangent
If $d y / d t=0$ and $d x / d t \neq 0$ when $t=t_0$, the curve represented by $x=f(t)$ and $y=g(t)$ has a horizontal tangent at $\left(f\left(t_0\right), g\left(t_0\right)\right)$.
Vertical Tangent
If $d x / d t=0$ and $d y / d t \neq 0$ when $t=t_0$, the curve represented by $x=f(t)$ and $y=g(t)$ has a vertical tangent at $\left(f\left(t_0\right), g\left(t_0\right)\right)$.
Example1
- A curve in the plane is defined parametrically by the equations $x=t^2$ and $y=t^3-2 t$.
(a) Sketch the curve in the $x y$-plane for $-2 \leq t \leq 2$. Indicate the direction in which the curve is traced as $t$ increases.
(b) For what values of $t$ does the curve have a vertical tangent?
(c) For what values of $t$ does the curve have a horizontal tangent?
(d) Find the equation of the tangent lines to the curve at $t= \pm \sqrt{2}$.
▶️Answer/Explanation
Solution
(a) Use a graphing calculator to draw a parametric curve.
Set the graphing calculator as follows.
Mode: Parametric mode
Window: $T_{\min }=-2, T_{\max }=2$, $X_{\min }=-5, X_{\max }=5$ $Y_{\min }=-4, Y_{\max }=4$
$Y=: \mathrm{X}_{1 \mathrm{~T}}=\mathrm{T}^2, \mathrm{Y}_{1 \mathrm{~T}}=\mathrm{T}^3-2 \mathrm{~T}$
(b) $\frac{d x}{d t}=2 t, \frac{d y}{d t}=3 t^2-2$
$
\frac{d x}{d t}=0 \Rightarrow t=0
$
The curve has a vertical tangent when $t=0$.
(c) $\frac{d y}{d t}=0 \Rightarrow 3 t^2-2=0 \Rightarrow t= \pm \frac{\sqrt{2}}{\sqrt{3}}= \pm \frac{\sqrt{6}}{3}$
The curve has a horizontal tangent when $t= \pm \frac{\sqrt{6}}{3}$
(d) If $t=\sqrt{2}, \frac{d y}{d x}=\frac{3(\sqrt{2})^2-2}{2(\sqrt{2})}=\frac{4}{2 \sqrt{2}}=\sqrt{2}$
The equation of the tangent line is
$
y-0=\sqrt{2}(x-2) \text {. }
$
If $t=-\sqrt{2}, \frac{d y}{d x}=\frac{3(-\sqrt{2})^2-2}{2(-\sqrt{2})}=\frac{4}{-2 \sqrt{2}}=-\sqrt{2}$
The equation of the tangent line is
$
y-0=-\sqrt{2}(x-2) \text {. }
$
Example 2
- A particle moves in the $x y$-plane so that its position at any time $t$, $0 \leq t \leq 4$, is given by the equations $x(t)=\cos t+t \sin t$ and $y(t)=\sin t-t \cos t$.
(a) Sketch the curve in the $x y$-plane for $0 \leq t \leq 4$. Indicate the direction in which the curve is traced as $t$ increases.
(b) At what time $t, 0<t<4$, does the line tangent to the path of the particle have a slope of -1 ?
(c) At what time $t, 0<t<4$, does $x(t)$ attain its maximum value? What is the position $(x(t), y(t))$ of the particle at this time?
(d) At what time $t, 0<t<4$, is the particle on the $y$-axis?
▶️Answer/Explanation
Solution
(a) Use a graphing calculator to draw the parametric curve.
o Mode: Parametric mode, Radian mode
Window: $T_{\min }=0, T_{\max }=4$, $X_{\min }=-5, X_{\max }=5$
$Y_{\min }=-4, Y_{\max }=4$
$Y=: X_{1 \mathrm{~T}}=\cos T+T \sin T$,
$Y_{1 \mathrm{~T}}=\sin T-T \cos T$
(b) $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{\cos t-(-t \sin t+\cos t)}{-\sin t+(t \cos t+\sin t)}=\frac{t \sin t}{t \cos t}=\tan t$
So $\frac{d y}{d x}=\tan t=-1 \Rightarrow t=\tan ^{-1}(-1)=3 \pi / 4$.
(c) $x^{\prime}(t)=-\sin t+(\sin t+t \cos t)=t \cos t$
$x^{\prime}(t)=0 \Rightarrow t=\pi / 2$, for $0<t<4$.
$x(t)$ attains its maximum value when $t=\pi / 2$.
$
\begin{aligned}
& x\left(\frac{\pi}{2}\right)=\cos \frac{\pi}{2}+\frac{\pi}{2} \sin \frac{\pi}{2}=\frac{\pi}{2} \\
& y\left(\frac{\pi}{2}\right)=\sin \frac{\pi}{2}-\frac{\pi}{2} \cos \frac{\pi}{2}=1
\end{aligned}
$
The position when $t=\pi / 2$ is $\left(\frac{\pi}{2}, 1\right)$.
(d) The particle is on the $y$-axis when $x(t)=0$.
$
x(t)=\cos t+t \sin t=0
$
Use a graphing calculator (in function mode) to find the value of $t$ which makes $\cos t+t \sin t=0$.
For $0<t<4, x(t)=0$ when $t=2.798$.
Example 3
If $x=t e^t$ and $y=t+e^t$, then $\frac{d y}{d x}$ at $t=0$ is
(A) 0
(B) $\frac{1}{2}$
(C) 1
(D) 2
▶️Answer/Explanation
Ans:D
Example 4
If $x=\tan t$ and $y=\sin t$, then $\frac{d^2 y}{d x^2}$ at $t=\frac{\pi}{6}$ is
(A) $-\frac{9}{11}$
(B) $-\frac{27}{32}$
(C) $\frac{13}{16}$
(D) $\frac{7}{8}$
▶️Answer/Explanation
Ans:B
Example 5
A curve $C$ is defined by the parametric equations $x=t^3-3$ and $y=2 t^2$. Which of the following is the equation for the line tangent to the graph of $C$ at the point $(5,8)$ ?
(A) $y=\frac{1}{3} x+\frac{8}{3}$
(B) $y=2 x-\frac{8}{3}$
(C) $y=\frac{2}{3} x+\frac{14}{3}$
(D) $y=3 x+8$
▶️Answer/Explanation
Ans:C
Example 6
If $x=\cos t$ and $y=2 \sin ^2 t$, then $\frac{d y}{d x}$ at $t=1$ is
(A) $-2 \cos 1$
(B) $-4 \cos 1$
(C) $-2 \tan 1$
(D) $-2 \sin 1$
▶️Answer/Explanation
Ans:B