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The chain rule gives $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ with nothing eliminated; the second derivative owes one more division by $dx/dt$; and the tangent is flat where the numerator vanishes and upright where the denominator does.
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 can differentiate a parametric curve without eliminating its parameter, using $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ and saying why the chain rule puts the $t$-derivative of $y$ on top. You can find the second derivative correctly, dividing by $dx/dt$ a second time rather than taking a ratio of second derivatives, and you can locate the parameters at which the tangent is horizontal or vertical.
You can differentiate anything in calculus AB, and you have met parametric equations in precalculus as a way of describing a path. This lesson does calculus on one, and the surprise is how little has to change: one chain rule, applied once, does all of it.
Parametric curve: $x$ and $y$ each given as a function of a parameter $t$.
Component derivatives: $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$, taken separately.
Slope: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$.
Horizontal tangent: where $\dfrac{dy}{dt} = 0$ and $\dfrac{dx}{dt}$ does not.
Vertical tangent: where $\dfrac{dx}{dt} = 0$ and $\dfrac{dy}{dt}$ does not.
A parametric curve gives $x$ and $y$ separately as functions of $t$. That lets it double back, cross itself, and be traced at any speed — none of which the graph of a function of $x$ can do.
The slope. The chain rule says $\dfrac{dy}{dt} = \dfrac{dy}{dx}\cdot\dfrac{dx}{dt}$. Divide, and
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} \qquad \text{wherever } \frac{dx}{dt} \neq 0.$$
Nothing is eliminated. For $x = t^2$, $y = t^3$ the slope is $\dfrac{3t^2}{2t} = \dfrac{3t}{2}$ in one line, and the Cartesian form $y^2 = x^3$ is never needed.
The second derivative is not a second ratio. $\dfrac{d^2y}{dx^2}$ is not $\dfrac{d^2y/dt^2}{d^2x/dt^2}$. It is the derivative of $\dfrac{dy}{dx}$ with respect to $x$, so
$$\frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{dx/dt}.$$
The same division by $dx/dt$ that produced the first derivative is owed again, because you are again converting a $t$-derivative into an $x$-derivative.
Flat and upright. The tangent is horizontal where the numerator $dy/dt$ vanishes and the denominator does not; vertical where the denominator vanishes and the numerator does not. Where both vanish the ratio is $0/0$ and the curve may have a cusp, which is a limit rather than a substitution.
Another way: picture
Picture a fly walking on a window while a camera photographs it every second. The parametric equations are the photographs; the Cartesian equation is the smear of ink left behind. The smear says where the fly went and nothing about when, how fast or which way — and the slope of the smear at a point is still recoverable from the photographs alone, by comparing how fast the two coordinates were changing.
Another way: steps
| Question | Answer | Note |
|---|---|---|
| Slope | $\dfrac{dy/dt}{dx/dt}$ | the $dt$ cancels |
| Bend | $\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right) \div \dfrac{dx}{dt}$ | one more division |
| Horizontal tangent | $\dfrac{dy}{dt} = 0$ | numerator vanishes |
| Vertical tangent | $\dfrac{dx}{dt} = 0$ | denominator vanishes |
Every row uses the same two derivatives. Deciding which question is being asked is the work; the algebra afterwards is short.
A graph $y = f(x)$ has one $y$ for each $x$, so it cannot be a circle, cannot cross itself and cannot double back. A parametric curve has none of those restrictions, and that is the reason for the extra letter.
$x = \cos t$, $y = \sin t$ traces the unit circle, and its slope is $\dfrac{\cos t}{-\sin t} = -\cot t$ — one expression covering the whole circle, top and bottom, with no $\pm$ and no two cases. Solving for $y$ instead gives $y = \pm\sqrt{1 - x^2}$, two functions, each differentiated separately, and a pair of points at $x = \pm 1$ where neither works.
The same curve can also be traced differently. $x = \cos 2t$, $y = \sin 2t$ draws the identical circle twice as fast, and every slope at every point comes out the same, because both component derivatives pick up the same factor of $2$ and the ratio does not notice. The picture is a property of the curve; the timing is a property of the parametrisation.
Eliminating the parameter first. It is unnecessary and usually impossible. $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ needs no Cartesian equation at all.
Taking the second derivative as a second ratio. $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ is not $\dfrac{d^2y}{dx^2}$ and is not anything else either. Differentiate $dy/dx$ with respect to $t$ and divide by $dx/dt$.
Substituting the value of $t$ too early. Put the number in after the division, not before. Before it, there is nothing left to differentiate again.
$x = t^2$, $y = t^3$: differentiate separately, giving $\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 3t^2$.
Two components, two derivatives.
$\dfrac{dy}{dx} = \dfrac{3t^2}{2t} = \dfrac{3t}{2}$.
The $dt$ cancels.
At $t = 2$ the slope is $3$ — and $y^2 = x^3$ was never written down.
Continuing from $\dfrac{dy}{dx} = \dfrac{3t}{2}$, differentiate that with respect to $t$: $\dfrac{3}{2}$.
Still a $t$-derivative.
Now divide by $\dfrac{dx}{dt} = 2t$: $\dfrac{3/2}{2t} = \dfrac{3}{4t}$.
The division that converts it.
Positive for $t > 0$, so the curve is concave up there. Stopping at $\tfrac32$ would have claimed a constant bend, which is false.
Differentiate separately: $\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 4$.
Components first.
Divide: $\dfrac{dy}{dx} = \dfrac{4}{2t} = \dfrac{2}{t}$.
Keep it as a function of $t$.
At $t = 2$ the slope is $1$ — and notice the tangent is vertical at $t = 0$, where the denominator vanishes.
Put the steps of finding $\dfrac{d^2y}{dx^2}$ for $x = t^2$, $y = 6t^3$ into the order you do them.
Number the steps in order (write the number in the box):
A curve has $x = e^{t}$ and $y = e^{2t}$. Which expression is $\dfrac{dy}{dx}$?
For $x = 2t^2$ and $y = 4t^3$, fill in the two component derivatives and their ratio at each value of $t$.
| $dx/dt$ | $dy/dt$ | $dy/dx$ | |
|---|---|---|---|
| $t = 1$ | |||
| $t = 2$ | |||
| $t = 3$ |
Match each question about a parametric curve to the combination of derivatives that answers it.
| $\dfrac{dy/dt}{dx/dt}$ | $\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right) \div \dfrac{dx}{dt}$ | $\dfrac{dy}{dt} = 0$ | $\dfrac{dx}{dt} = 0$ | |
|---|---|---|---|---|
| The slope of the curve | ||||
| How the curve bends | ||||
| Where the tangent is horizontal | ||||
| Where the tangent is vertical |
A curve has $x = 6t^2$ and $y = 3t^3$. Find $\dfrac{dy}{dx}$ at $t = 3$.
Answer:
A curve has $x = t^2 - 4t$ and $y = t^3 - 27t$. For $t > 0$, give the value of $t$ at which the tangent is horizontal, then the value at which it is vertical.
horizontal at t = h, vertical at t = v
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the steps of finding $\dfrac{d^2y}{dx^2}$ for $x = t^2$, $y = 4t^3$ into the order you do them.
Number the steps in order (write the number in the box):
You can differentiate a parametric curve without eliminating anything. Without looking: what is $\dfrac{dy}{dx}$ in terms of the two $t$-derivatives, and what is the extra step the second derivative needs?
10. Your turn: the slope of $x = t^2 + 1$, $y = 4t$ at $t = 2$, step 3