Exercises — Parametric differentiation — dy - dx, d²y - dx²

What we do: differentiate each coordinate separately with respect to $t$.

• $\dot{x}=\frac{d}{dt}(3t+1)=3$ — the $3t$ has slope $3$, the constant $1$ contributes nothing.
• $\dot{y}=\frac{d}{dt}(t^2-4)=2t$ — power rule; the $-4$ vanishes.

Why divide: slope is vertical-speed over horizontal-speed (the shared $dt$ cancels): $\frac{dy}{dx}=\frac{\dot{y}}{\dot{x}}=\frac{2t}{3}.$

$\frac{d^{2}y}{d{x}^2}$ means "differentiate $\frac{dy}{dx}$ with respect to $x$, not $t$," so you must re-apply the parametric trick to $\frac{\dot{y}}{\dot{x}}$, producing a quotient-rule numerator over a denominator of $=={\dot{x}}^3==$ — the cube, not $\ddot{x}$.

• $\dot{x}=2t$, $\dot{y}=3t^2$.
• $\frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2}$ (valid for $t\ne 0$, where $\dot{x}\ne 0$).
• At $t=2$: $\frac{dy}{dx}=\frac{3\cdot 2}{2}=3.$

Tool: the boxed second-derivative formula. We already have $\dot{x}=2t,\dot{y}=3t^2$. Now:

• $\ddot{x}=2$, $\ddot{y}=6t$.
• Plug in: $\frac{d^{2}y}{d{x}^2}=\frac{\ddot{y}\dot{x}-\dot{y}\ddot{x}}{{\dot{x}}^3}=\frac{(6t)(2t)-(3t^2)(2)}{(2t{)}^3}=\frac{12t^2-6t^2}{8t^3}=\frac{6t^2}{8t^3}=\frac{3}{4t}.$ Cross-check via the "differentiate $dy/dx$ then divide" route: $\frac{dy}{dx}=\frac{3t}{2}$, so $\frac{d}{dt}\left(\frac{3t}{2}\right)=\frac{3}{2}$; divide by $\dot{x}=2t$: $\frac{3/2}{2t}=\frac{3}{4t}$. ✓ Same answer.

• $\dot{x}=-R\sin t$, $\dot{y}=R\cos t$.
• $\frac{dy}{dx}=\frac{R\cos t}{-R\sin t}=-\cot t.$ The $R$ cancels — a bigger circle has the same slope at the same angle.
• At $t=\pi /6$: $-\cot (\pi /6)=-\sqrt{3}\approx -\mathrm{1.732.}$ (Independent of $R$, so $R=2$ doesn't change it.)

Horizontal tangent = slope $0$ = "vertical speed is zero while horizontal speed isn't": $\dot{y}=0,\dot{x}\ne 0$.

• $\dot{y}=\cos t=0\Rightarrow t=\frac{\pi }{2},\frac{3\pi }{2}$. At both, $\dot{x}=-\sin t\ne 0$. These are the top $(0,1)$ and bottom $(0,-1)$ — look at the red horizontal arrows in the figure.

Vertical tangent = run is zero = "horizontal speed is zero while vertical isn't": $\dot{x}=0,\dot{y}\ne 0$.

• $\dot{x}=-\sin t=0\Rightarrow t=0,\pi$. At both, $\dot{y}=\cos t\ne 0$. These are the right $(1,0)$ and left $(-1,0)$ points, where $\frac{dy}{dx}$ is undefined — not an error, just geometry.

• $\dot{x}=1,\dot{y}=3t^2$; $\ddot{x}=0,\ddot{y}=6t$.
• $\frac{d^{2}y}{d{x}^2}=\frac{(6t)(1)-(3t^2)(0)}{1^3}=6t=6x.$

Case analysis (sign of $6t$ decides — see Concavity and second derivative):

• $t>0$: positive → concave up (bowl opens upward).
• $t<0$: negative → concave down.
• $t=0$: zero → an inflection point where concavity flips, at the origin.

• $\dot{x}=1-\cos t,\dot{y}=\sin t$, so $\frac{dy}{dx}=\frac{\sin t}{1-\cos t}$.
• At $t=\pi /2$: $\frac{\sin (\pi /2)}{1-\cos (\pi /2)}=\frac{1}{1-0}=1.$
• Limit $t\to 0^{+}$: both top and bottom $\to 0$ ($\frac{0}{0}$). Using the small-angle pictures, $\sin t\approx t$ and $1-\cos t\approx \frac{t^2}{2}$, so $\frac{dy}{dx}\approx \frac{t}{t^2/2}=\frac{2}{t}\to +\infty$. The cycloid rises with a vertical tangent at each cusp (where the wheel touches the ground) — the curve comes into the cusp straight up.

Step 1 — the point. $x=2\cos \frac{\pi }{4}=2\cdot \frac{\sqrt{2}}{2}=\sqrt{2}$; $y=3\sin \frac{\pi }{4}=\frac{3\sqrt{2}}{2}$. Step 2 — the slope (see Tangents and normals).

• $\dot{x}=-2\sin t,\dot{y}=3\cos t$, so $\frac{dy}{dx}=\frac{3\cos t}{-2\sin t}=-\frac{3}{2}\cot t$.
• At $t=\pi /4$, $\cot (\pi /4)=1$, so $m=-\frac{3}{2}$. Step 3 — the line through $(\sqrt{2},\frac{3\sqrt{2}}{2})$ with slope $-\frac{3}{2}$: $y-\frac{3\sqrt{2}}{2}=-\frac{3}{2}\left(x-\sqrt{2}\right)\Rightarrow y=-\frac{3}{2}x+\frac{3}{2}\sqrt{2}+\frac{3\sqrt{2}}{2}=-\frac{3}{2}x+3\sqrt{2}.$ So $m=-\frac{3}{2},c=3\sqrt{2}\approx 4.243$.

• $\dot{x}=-2\sin t,\dot{y}=3\cos t$; $\ddot{x}=-2\cos t,\ddot{y}=-3\sin t$.
• Numerator $\ddot{y}\dot{x}-\dot{y}\ddot{x}=(-3\sin t)(-2\sin t)-(3\cos t)(-2\cos t)=6{\sin }^{2}t+6{\cos }^{2}t=6.$ (Used ${\sin }^{2}t+{\cos }^{2}t=1$.)
• Denominator ${\dot{x}}^3=(-2\sin t{)}^3=-8{\sin }^{3}t.$
• Therefore $\frac{d^{2}y}{d{x}^2}=\frac{6}{-8{\sin }^{3}t}=-\frac{3}{4{\sin }^{3}t}.$ $■$

• $\dot{x}=2t,\dot{y}=3t^2-3=3(t^2-1)$.
• (a) Horizontal ($\dot{y}=0,\dot{x}\ne 0$): $3(t^2-1)=0\Rightarrow t=\pm 1$; both have $\dot{x}=\pm 2\ne 0$. ✓
• (b) Vertical ($\dot{x}=0,\dot{y}\ne 0$): $2t=0\Rightarrow t=0$; there $\dot{y}=-3\ne 0$. ✓ Vertical tangent at the origin-side point $(0,0)$.
• (c) Concavity at $t=1$. Need $\frac{d^{2}y}{d{x}^2}$: $\ddot{x}=2,\ddot{y}=6t$. $\frac{d^{2}y}{d{x}^2}=\frac{(6t)(2t)-3(t^2-1)(2)}{(2t{)}^3}=\frac{12t^2-6t^2+6}{8t^3}=\frac{6t^2+6}{8t^3}=\frac{3(t^2+1)}{4t^3}.$ At $t=1$: $\frac{3(2)}{4}=\frac{3}{2}>0$ → concave up.

Step 1 — restate the target. $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$ — differentiate the slope with respect to $x$. Step 2 — we only know $t$-derivatives, so apply the parametric trick to the function $g=\frac{dy}{dx}=\frac{\dot{y}}{\dot{x}}$: $\frac{d}{dx}(g)=\frac{dg/dt}{dx/dt}=\frac{1}{\dot{x}}\cdot \frac{d}{dt}\left(\frac{\dot{y}}{\dot{x}}\right).$ Step 3 — quotient rule on $\frac{\dot{y}}{\dot{x}}$ (top' · bottom − top · bottom', over bottom²): $\frac{d}{dt}\left(\frac{\dot{y}}{\dot{x}}\right)=\frac{\ddot{y}\dot{x}-\dot{y}\ddot{x}}{{\dot{x}}^2}.$ Step 4 — combine (multiply by the $\frac{1}{\dot{x}}$ from Step 2): $\frac{d^{2}y}{d{x}^2}=\frac{1}{\dot{x}}\cdot \frac{\ddot{y}\dot{x}-\dot{y}\ddot{x}}{{\dot{x}}^2}=\frac{\ddot{y}\dot{x}-\dot{y}\ddot{x}}{{\dot{x}}^3}.■$ The ${\dot{x}}^2$ from the quotient rule times the extra $\dot{x}$ from Step 2 is exactly where the cube is born.

Note the curve is $y=(e^t{)}^2=x^2$, so we know the true answer must be $\frac{d^{2}y}{d{x}^2}=2$.

• $\dot{x}=e^t,\dot{y}=2e^{2t}$; $\ddot{x}=e^t,\ddot{y}=4e^{2t}$.
• (a) Correct: $\frac{\ddot{y}\dot{x}-\dot{y}\ddot{x}}{{\dot{x}}^3}=\frac{(4e^{2t})(e^t)-(2e^{2t})(e^t)}{(e^t{)}^3}=\frac{4e^{3t}-2e^{3t}}{e^{3t}}=\frac{2e^{3t}}{e^{3t}}=2.$ ✓ (matches $y=x^2$).
• (b) Naive wrong: $\frac{\ddot{y}}{\ddot{x}}=\frac{4e^{2t}}{e^t}=4e^t=4x.$
• (c) Disagreement: the correct answer is the constant $2$; the naive one is $4x$, which even changes with $t$. They coincide nowhere except $4x=2\Rightarrow x=\frac{1}{2}$, a coincidence at a single point — proof the naive formula is genuinely wrong.

Original ($x=t,y=t^2$): $\dot{x}=1,\dot{y}=2t$, so $\frac{dy}{dx}=2t=2x.$ Reparametrized ($x=u^3,y=u^6$): here dots mean $d/du$.

• $\dot{x}=3u^2,\dot{y}=6u^5$.
• $\frac{dy}{dx}=\frac{6u^5}{3u^2}=2u^3.$
• But $x=u^3$, so $2u^3=2x.$ Identical. ✓ The parameter is a scaffold; the curve and its slope are the real objects. This is the deep reason the shared $dt$ (or $du$) cancels — a lesson that carries into all Parametric curves.