Instead of one equation, you get two: \(x(t)\) and \(y(t)\). As t advances, the point \((x(t),\,y(t))\) moves, and the trail it leaves is the curve. Drag the slider to move time forward, or hit play. Switch curves to see why one function of x could never capture these.
The dashed cyan line is the tangent — its slope is exactly \(\tfrac{dy}{dx}\), which we're about to compute. Notice it tilts and even goes vertical as the point moves; that's the first hint that slope in parametric land needs its own formula.
Before doing any calculus, it helps to recognize the curve. If you can get rid of t and write a plain relationship between x and y, you often discover the parametric pair is just a familiar shape — a line, a parabola, a circle — in disguise.
When one equation is easy to solve for t, do that and plug into the other. Take \(x=t+1\), \(y=t^2\). Solve the first: \(t=x-1\). Substitute into the second:
So this parametric pair traces the parabola \(y=(x-1)^2\) — same curve you'd graph in Algebra II, now just "clocked" by t.
When you see cos and sin, reach for \(\cos^2\theta+\sin^2\theta=1\). Take \(x=3\cos t\), \(y=3\sin t\). Then \(\cos t=\tfrac{x}{3}\) and \(\sin t=\tfrac{y}{3}\), so:
That's a circle of radius 3 centered at the origin. (If the coefficients differ — \(x=2\cos t\), \(y=3\sin t\) — the same trick gives \(\left(\tfrac{x}{2}\right)^2+\left(\tfrac{y}{3}\right)^2=1\), an ellipse.)
The rectangular equation may describe more of the curve than the parametric form actually traces. Example: \(x=t^2,\ y=t\) gives \(x=y^2\) — but since \(x=t^2\ge 0\) always, the parametric curve is only the right half (\(x\ge 0\)) of that sideways parabola. Always check what range of \(x\) and \(y\) the parameter really produces.
You want \(\tfrac{dy}{dx}\), but you only have x and y in terms of t. The chain rule bridges them: \(\tfrac{dy}{dt}=\tfrac{dy}{dx}\cdot\tfrac{dx}{dt}\). Solve for what you want:
In words: the rate y changes with t, divided by the rate x changes with t. The t's "cancel" the way fractions suggest. This is the slope of the tangent line to the curve at the point for that t.
The tangent is horizontal where \(\tfrac{dy}{dt}=0\) (and \(\tfrac{dx}{dt}\neq 0\)) — the curve momentarily stops rising. It's vertical where \(\tfrac{dx}{dt}=0\) (and \(\tfrac{dy}{dt}\neq 0\)) — the slope blows up. AP loves asking for both.
Here's where most points are lost. To get \(\tfrac{d^2y}{dx^2}\) you do not just take \(\tfrac{d^2y}{dt^2}\). The second derivative is the rate of change of the first derivative with respect to x — so you apply the exact same parametric rule again, this time to \(\tfrac{dy}{dx}\):
Read it as a recipe: take the derivative of \(\tfrac{dy}{dx}\) with respect to t, then divide by \(\tfrac{dx}{dt}\) again. The denominator is \(\tfrac{dx}{dt}\), never dt or dx alone.
\(\dfrac{d^2y}{dx^2}\) is NOT \(\dfrac{d^2y}{dt^2}\div\dfrac{d^2x}{dt^2}\). You must differentiate the slope \(\tfrac{dy}{dx}\) with respect to \(t\), then divide by \(\tfrac{dx}{dt}\). Forgetting that final "\(\div\,dx/dt\)" is the single most common error on this topic.
In Unit 8 you found arc length for \(y=f(x)\) by adding up tiny hypotenuses \(\sqrt{dx^2+dy^2}\), factoring out dx. Parametric arc length is the same idea — but now both x and y change with t, so factor out dt instead:
Each little step along the curve has horizontal part \((dx/dt)\,dt\) and vertical part \((dy/dt)\,dt\); the hypotenuse is \(\sqrt{(dx/dt)^2+(dy/dt)^2}\,dt\), and the integral sums them from \(t=a\) to \(t=b\). Same Pythagorean theorem as Unit 8 — it just has two moving legs now.
Both derivatives get squared under the root: \((dx/dt)^2\) and \((dy/dt)^2\). And the bounds are values of t, not x — make sure your limits match the parameter.
For \(x=t^2\), \(y=t^3\), find \(\dfrac{dy}{dx}\) and \(\dfrac{d^2y}{dx^2}\) at \(t=2\).
First derivative. So at \(t=2\), \(\dfrac{dy}{dx}=\dfrac{3(2)}{2}=3\). dy/dx = 3
Second derivative. Differentiate \(\dfrac{dy}{dx}=\dfrac{3t}{2}\) with respect to \(t\) (that's \(\tfrac{3}{2}\)), then divide by \(\dfrac{dx}{dt}=2t\): At \(t=2\): \(\dfrac{3}{4\cdot 2}=\dfrac{3}{8}\). d²y/dx² = 3/8
For \(x=t^2+1\), \(y=t^3-3t\), where are the tangents horizontal or vertical?
Horizontal: set \(\dfrac{dy}{dt}=3t^2-3=0\Rightarrow t=\pm 1\). Points: \((2,-2)\) at \(t=1\) and \((2,2)\) at \(t=-1\). horizontal at t = ±1
Vertical: set \(\dfrac{dx}{dt}=2t=0\Rightarrow t=0\). Point: \((1,0)\). (Check \(\dfrac{dy}{dt}=-3\neq 0\) there, so it's genuinely vertical.) vertical at t = 0
Find the length of \(x=t^2\), \(y=t^3\) from \(t=0\) to \(t=1\).
Set up. \(\dfrac{dx}{dt}=2t\), \(\dfrac{dy}{dt}=3t^2\), so
Simplify & integrate. Factor: \(\sqrt{4t^2+9t^4}=t\sqrt{4+9t^2}\) (since \(t\ge 0\)). Substitute \(u=4+9t^2\), \(du=18t\,dt\):
Result. \(=\tfrac{1}{27}\big(13^{3/2}-8\big)\approx 1.44\). L = (13√13 − 8)/27 ≈ 1.44
Five questions on slope, concavity, tangents, and arc length. Pick the right setup or value.