The Projectile's Path
A ball is launched from ground level so its position at time $t$ seconds is $x=5t$ and $y=20t-5t^{2}$. The trajectory is a parabola, but $y$ is not given as a function of $x$ — both coordinates are driven by a hidden clock $t$. What is the slope of the path?
$\dfrac{dx}{dt}$ = 5.0 · $\dfrac{dy}{dt}$ = 10.0 · $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ = 2.00
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. The chain rule links rates as $\dfrac{dy}{dx}=$ ?
R2. $\dfrac{d}{dt}\big(t^{3}\big)$ = ?
R3. $\dfrac{d}{dt}\big(e^{2t}\big)$ = ?
Both $x(t)$ and $y(t)$ are functions of $t$. The chain rule writes the slope of the curve as a quotient of two $t$-rates:
Differentiate $x$ and $y$ separately with respect to $t$, then divide. No need to eliminate the parameter.
| Parametric curve | $\dfrac{dy}{dx}$ |
|---|---|
| $x=3t^2,\ y=2t^3$ | $\dfrac{6t^2}{6t}=t$ |
| $x=t^2,\ y=t^3$ | $\dfrac{3t^2}{2t}=\dfrac{3t}{2}$ |
| $x=3\cos\theta,\ y=2\sin\theta$ | $-\dfrac{2}{3}\cot\theta$ |
| $x=e^{2t}+1,\ y=e^t$ | $\dfrac{1}{2e^t}$ |
Simplify the quotient whenever possible — powers of $t$ often cancel.
$\dfrac{dx}{dt}=5$ and $\dfrac{dy}{dt}=20-10t$, so $\dfrac{dy}{dx}=\dfrac{20-10t}{5}=4-2t$.
At $t=1$: $\dfrac{dy}{dx}=4-2=2$ — the path is rising at slope $2$, matching the tangent you dragged in the hook.
Groups of 4: tap a parametric pair on the left, then its dy/dx on the right — a coloured line links each pair. Six curves covering polynomial, trig and exponential parameters. Wrong pairs turn red to fix.
Each attempt below claims a dy/dx. Decide whether the parametric method was applied correctly, then explain your verdict out loud to a partner.
The curve $x=at^2,\ y=2at$ has parameter $t$. Find $\dfrac{dy}{dx}$ and the equation of the tangent at the point with parameter $t$.
$\dfrac{dx}{dt}=2at,\ \dfrac{dy}{dt}=2a$, so $\dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}$.
$y-2at=\dfrac{1}{t}(x-at^2)\ \Rightarrow\ y=\dfrac{x}{t}+at$.
The curve $x=e^{3t},\ y=t^{2}e^{t}+3$. Show that $\dfrac{dy}{dx}=\dfrac{t(t+2)}{3e^{2t}}$.
$\dfrac{dx}{dt}=3e^{3t}$ and $\dfrac{dy}{dt}=2te^{t}+t^{2}e^{t}=e^{t}(2t+t^{2})=te^{t}(t+2)$.
$\dfrac{dy}{dx}=\dfrac{te^{t}(t+2)}{3e^{3t}}=\dfrac{t(t+2)}{3e^{2t}}$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. For $x=t^{2},\ y=t^{3}$, $\dfrac{dy}{dx}$ = ?
Q2. For $x=2t,\ y=t^{2}$, the gradient at $t=3$ is:
Q3. For $x=3\cos\theta,\ y=2\sin\theta$, $\dfrac{dy}{dx}$ = ?
Q4. The tangent to $x=t^{2},\ y=t^{3}$ at $t=1$ is:
Work each on paper, then reveal the solution to check your working.
Find $\dfrac{dy}{dx}$ for $x=\theta-\cos\theta,\ y=\theta+\sin\theta$.
Reveal Solution
$\dfrac{dx}{d\theta}=1+\sin\theta$ and $\dfrac{dy}{d\theta}=1+\cos\theta$, so $\dfrac{dy}{dx}=\dfrac{1+\cos\theta}{1+\sin\theta}$.
Find $\dfrac{dy}{dx}$ for $x=t+\dfrac{1}{t},\ y=t-\dfrac{1}{t}$.
Reveal Solution
$\dfrac{dx}{dt}=1-t^{-2},\ \dfrac{dy}{dt}=1+t^{-2}$, so $\dfrac{dy}{dx}=\dfrac{1+t^{-2}}{1-t^{-2}}=\dfrac{t^2+1}{t^2-1}$.
Find $\dfrac{dy}{dx}$ for $x=(t+1)^2,\ y=(t-1)^2$.
Reveal Solution
$\dfrac{dx}{dt}=2(t+1),\ \dfrac{dy}{dt}=2(t-1)$, so $\dfrac{dy}{dx}=\dfrac{2(t-1)}{2(t+1)}=\dfrac{t-1}{t+1}$.
Find $\dfrac{dy}{dx}$ for $x=\dfrac{t}{1+t},\ y=\dfrac{t}{1-t}$.
Reveal Solution
$\dfrac{dx}{dt}=\dfrac{1}{(1+t)^2}$ and $\dfrac{dy}{dt}=\dfrac{1}{(1-t)^2}$, so $\dfrac{dy}{dx}=\dfrac{(1+t)^2}{(1-t)^2}$.
The curve $x=t^2,\ y=1-\dfrac{1}{2t}$ (for $t>0$) cuts the $x$-axis at P. Find the coordinates of P, the gradient there, and the equation of the tangent at P.
Reveal Solution
$y=0$ gives $t=\tfrac12$, so $x=\tfrac14$ and $P=\left(\tfrac14,0\right)$.
$\dfrac{dy}{dx}=\dfrac{1/(2t^2)}{2t}=\dfrac{1}{4t^3}$, so at $t=\tfrac12$ the gradient is $2$.
Tangent: $y-0=2\left(x-\tfrac14\right)$, i.e. $y=2x-\tfrac12$, which cuts the $y$-axis at $\left(0,-\tfrac12\right)$.
Exit Ticket
Check the learning objectives you feel confident about: