Back to the on-screen lesson ·
A curve given by where a point is at each time; the slope is a ratio of parameter-derivatives and the length is the speed integrated over time.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to differentiate each coordinate with respect to the parameter and form the slope as their ratio, find horizontal and vertical tangents by setting the right derivative to zero and checking the other, compute the second derivative by differentiating the slope with respect to the parameter and dividing by $x'(t)$, calculate speed as a magnitude and arc length as the integral of speed over a range that traces the curve once, and say which quantities belong to the path and which to the parametrisation.
Everything so far has been $y = f(x)$: one height for each horizontal position. A circle is not of that form, and neither is any curve that doubles back or crosses itself.
The fix is to stop describing the curve by a relation between $x$ and $y$ and start describing it by where a point is at each time: $x = x(t)$, $y = y(t)$. That covers every curve a point can trace, and it carries extra information — the timing — which the picture does not show. Separating what is a property of the path from what is a property of the timing is most of what this lesson is about.
The parameter $t$ is the input; it is often time but need not be. A parametrisation is a particular pair of functions describing a curve, and one curve has many.
Velocity is the pair $(x'(t), y'(t))$; speed is its magnitude $\sqrt{x'^2 + y'^2}$, a single number. Speed depends on the parametrisation; the shape of the path does not.
A parametrisation is smooth where $x'$ and $y'$ are continuous and not both zero. Where both vanish the curve can have a corner or a cusp even though the coordinate functions are perfectly differentiable, which is why that condition is part of the definition.
Slope. Where $x'(t) \ne 0$, the chain rule gives $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{y'(t)}{x'(t)}.$$ The $dt$ cancels, so the slope is a property of the path and not of the timing.
Vertical and horizontal tangents. Horizontal where $y' = 0$ and $x' \ne 0$; vertical where $x' = 0$ and $y' \ne 0$. Both zero is a singular point — a cusp or a corner — and the slope there needs a limit rather than a substitution.
Second derivative. $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{x'(t)}.$$ Note the division by $x'(t)$ and not by $x''(t)$ — differentiating with respect to $x$ always means differentiating with respect to $t$ and dividing.
Speed and arc length. Speed is $\sqrt{x'^2 + y'^2}$, and $$L = \int_\alpha^\beta \sqrt{x'(t)^2 + y'(t)^2}\,dt$$ — the speed integrated over time, which is why it is a distance. Setting $x = t$ recovers $\int\sqrt{1 + (y')^2}\,dx$ from lesson 11, so this is the same formula in a more general notation.
Area under a curve. $\displaystyle\int y\,dx = \int_\alpha^\beta y(t)\,x'(t)\,dt$ — the substitution rule, applied to the area integral.
One trace only. The length formula counts every pass, so a parametrisation that traverses a curve twice gives twice the length. Choosing the parameter range that traces the curve exactly once is part of setting the problem up.
Another way: picture
A particle moving along a curve with its velocity drawn as an arrow at each instant: the arrow points along the direction of travel and its length is the speed. The path is the set of positions, the arrow is the derivative, and the slope of the path is the arrow's direction — which is unchanged if the particle speeds up, and reversed if it turns round.
Another way: steps
The unit circle can be traced as $(\cos t, \sin t)$ for $t \in [0, 2\pi]$, or as $(\cos 2t, \sin 2t)$ for $t \in [0, \pi]$, or as $(\cos t, -\sin t)$, or as $(\cos t^2, \sin t^2)$. Same picture, four different parametrisations.
What they agree on: the set of points, the slope at each point, the total length $2\pi$, and the area enclosed. These are properties of the path.
What they disagree on: the velocity at each instant, the speed, how long the trace takes, and the direction of travel. These are properties of the parametrisation.
The test is whether $t$ survives into the answer. $\frac{dy}{dx} = \frac{y'}{x'}$ has the $dt$ cancelled — a path property. Speed is $\sqrt{x'^2 + y'^2}$ with nothing cancelled — a timing property. Arc length looks like a timing quantity and is not, because $\sqrt{x'^2 + y'^2}\,dt$ is invariant under reparametrisation: doubling the speed halves the time.
This is why the length formula requires the range to trace the curve exactly once. $(\cos t, \sin t)$ on $[0, 4\pi]$ gives $4\pi$, and that is the distance the point travelled — which is a correct answer to a different question from the length of the circle.
Dividing the second derivatives. $\dfrac{d^2y}{dx^2} \ne \dfrac{y''(t)}{x''(t)}$. Differentiate the slope with respect to $t$ and divide by $x'(t)$.
Substituting the parameter value before differentiating. Then there is nothing left to differentiate.
Adding the velocity components to get speed. They combine at right angles.
Missing a vertical tangent. A graph has none, so the habit of setting only the numerator to zero survives. Set the denominator to zero as well.
Reading both derivatives vanishing as an ordinary point. That is a singular point, possibly a cusp, and the slope there is a limit.
Integrating over a range that traces the curve twice. The length comes out twice as large, correctly, for a different question.
Forgetting to change $dx$ in an area integral. $\int y\,dx$ becomes $\int y(t)x'(t)\,dt$; the $x'(t)$ is not optional.
Because $t$ is written alongside $x$ and $y$, it is easy to treat it as another axis and to expect it to appear in the answers. It does not appear in any of them, and the reason is worth holding onto: $t$ is an index on the points, not a location.
The clearest symptom is writing $\dfrac{dy}{dx} = \dfrac{y''(t)}{x''(t)}$ for the second derivative, by analogy with the first. The first derivative's ratio worked because the $dt$ cancelled between a numerator and a denominator that were both first-order in $dt$. Second derivatives are not in that relationship, and the correct formula divides $\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)$ by $x'(t)$ — one $t$-derivative, one division, exactly as before.
The other symptom is treating the drawn curve as the whole object. Two parametrisations can draw the same picture and disagree about everything to do with motion. Speed is not visible on the picture; slope is. Length is a property of the picture only if the parameter range traces it once.
So when an answer comes out with a $t$ in it and the question was geometric, something has not cancelled that should have — and when a question asks for speed and the answer has no $t$, the curve is probably being traced at a constant rate, which is worth checking rather than assuming.
$x = t - \sin t$, $y = 1 - \cos t$ — the path of a point on a rolling wheel. $\dfrac{dx}{dt} = 1 - \cos t$ and $\dfrac{dy}{dt} = \sin t$, so $\dfrac{dy}{dx} = \dfrac{\sin t}{1 - \cos t}$.
Both derivatives, then the ratio.
At $t = \pi/2$ the slope is $\dfrac{1}{1} = 1$. At $t = 0$ both derivatives vanish: a singular point, where the wheel's marked point touches the ground.
A point the formula cannot be substituted into.
Length of one arch, $0 \le t \le 2\pi$: the speed is $\sqrt{(1-\cos t)^2 + \sin^2 t} = \sqrt{2 - 2\cos t} = 2\left|\sin\tfrac t2\right|$, so $L = \int_0^{2\pi}2\sin\tfrac t2\,dt = 8$. Exactly eight times the wheel's radius, with no $\pi$ in it at all.
$x = t^2 - 1$, $y = t^3 - t$. At $t = 1$ and $t = -1$ both give the point $(0, 0)$: the curve passes through the origin twice.
Not a graph, and not describable as one.
$\dfrac{dy}{dx} = \dfrac{3t^2 - 1}{2t}$. At $t = 1$ that is $1$; at $t = -1$ it is $-1$.
One point, two tangent lines.
So the curve has two different slopes at the same point, which is impossible for a graph and ordinary for a parametric curve. The parameter is what distinguishes the two passes.
$\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 3t^{2} - 3$.
Both derivatives first.
Horizontal where $\dfrac{dy}{dt} = 0$ and $\dfrac{dx}{dt} \ne 0$: at $t = \pm1$, where $x' = \pm2 \ne 0$. Two horizontal tangents.
Numerator zero, denominator not.
Vertical where $\dfrac{dx}{dt} = 0$ and $\dfrac{dy}{dt} \ne 0$: at $t = 0$, where $y' = -3 \ne 0$. So the curve is vertical at the origin, which no graph could be.
A curve is $x = 3t$, $y = 3t^{2}$. Find $\dfrac{dy}{dx}$ at $t = 4$.
Answer:
A point moves with $x = 15t$ and $y = 8t$. What is its speed?
Answer:
Plot the points of the curve $x = t$, $y = t^{2}$ at $t = 1$, $t = 2$ and $t = 3$.
Plot your answer on the grid:
Put the steps of finding the tangent line to $x = t^{2}$, $y = 2t^{3}$ at a given $t$ into the order you do them.
Number the steps in order (write the number in the box):
Match each quantity to its formula for a curve $x = x(t)$, $y = y(t)$.
| $\dfrac{y'(t)}{x'(t)}$ | $\sqrt{x'(t)^{2} + y'(t)^{2}}$ | $\displaystyle\int\sqrt{x'(t)^{2} + y'(t)^{2}}\,dt$ | $\displaystyle\int y(t)\,x'(t)\,dt$ | |
|---|---|---|---|---|
| The slope $dy/dx$ | ||||
| The speed | ||||
| The arc length | ||||
| The area under the curve |
A curve is $x = 6t^{2} - 3t$, $y = t^{3} + t$. At which $t$ is the tangent vertical?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $x = 24t$, $y = 7t$, find the arc length from $t = 0$ to $t = 1$.
Answer:
You can do calculus on a curve that is not a graph, and say what changes when the same path is traced at a different speed. Say in your own words why $d^2y/dx^2$ is not the ratio of the second derivatives. Next: describing a curve by how far out it reaches at each angle.
10. Your turn: $x = t^{2}$, $y = t^{3} - 3t$. Where is the tangent horizontal, and where vertical?, step 3
Denominator zero, numerator not.