provided $f'(t) \neq 0$.

💡 Key Idea: Use the chain rule! Since both $x$ and $y$ depend on $t$, divide the rates of change.

Why This Formula Works

By the chain rule: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$

Solving for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Think: "How fast is $y$ changing compared to how fast $x$ is changing?"

Example 1: Finding dy/dx

For $x = t^2$, $y = t^3$, find $\frac{dy}{dx}$.

Step 1: Find derivatives with respect to $t$

$\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2$

Step 2: Apply formula

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2t} = \frac{3t}{2}$

Step 3: Express in terms of $x$ (optional)

Since $x = t^2$, we have $t = \pm\sqrt{x}$

$\frac{dy}{dx} = \frac{3\sqrt{x}}{2} \text{ or } \frac{-3\sqrt{x}}{2}$

(depending on which branch)

Example 2: Slope at a Point

For $x = 3\cos t$, $y = 2\sin t$, find the slope at $t = \frac{\pi}{4}$.

Step 1: Find derivatives

$\frac{dx}{dt} = -3\sin t, \quad \frac{dy}{dt} = 2\cos t$

Step 2: Find dy/dx

$\frac{dy}{dx} = \frac{2\cos t}{-3\sin t} = -\frac{2\cos t}{3\sin t} = -\frac{2}{3}\cot t$

Step 3: Evaluate at $t = \frac{\pi}{4}$

$\frac{dy}{dx}\bigg|_{t=\pi/4} = -\frac{2}{3}\cot\frac{\pi}{4} = -\frac{2}{3}(1) = -\frac{2}{3}$

Answer: The slope at $t = \frac{\pi}{4}$ is $-\frac{2}{3}$.

Horizontal and Vertical Tangents

Horizontal Tangent

Occurs when $\frac{dy}{dx} = 0$

This happens when:

• $\frac{dy}{dt} = 0$ AND
• $\frac{dx}{dt} \neq 0$

Vertical Tangent

Occurs when $\frac{dy}{dx}$ is undefined

This happens when:

• $\frac{dx}{dt} = 0$ AND
• $\frac{dy}{dt} \neq 0$

Singular Point

If $\frac{dx}{dt} = 0$ AND $\frac{dy}{dt} = 0$ simultaneously:

• The curve might have a cusp, self-intersection, or other singularity
• Need more analysis (possibly using higher derivatives)

Example 3: Finding Tangent Lines

For $x = t^3 - 3t$, $y = t^2$, find all points where the tangent is horizontal.

Step 1: Find derivatives

$\frac{dx}{dt} = 3t^2 - 3, \quad \frac{dy}{dt} = 2t$

Step 2: Set $\frac{dy}{dt} = 0$

$2t = 0 \implies t = 0$

Step 3: Check that $\frac{dx}{dt} \neq 0$ at $t = 0$

$\frac{dx}{dt}\bigg|_{t=0} = 3(0)^2 - 3 = -3 \neq 0$ ✓

Step 4: Find the point

At $t = 0$: $x = 0^3 - 3(0) = 0$ $y = 0^2 = 0$

Answer: Horizontal tangent at point $(0, 0)$.

Second Derivative

To find $\frac{d^2y}{dx^2}$ for parametric curves:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d/dt(dy/dx)}{dx/dt}$

Process:

1. Find $\frac{dy}{dx}$ as before
2. Take derivative with respect to $t$: $\frac{d}{dt}\left(\frac{dy}{dx}\right)$
3. Divide by $\frac{dx}{dt}$

Example 4: Second Derivative

For $x = t^2$, $y = t^3$, find $\frac{d^2y}{dx^2}$.

Step 1: Find first derivative

$\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$

Step 2: Differentiate with respect to $t$

$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{3t}{2}\right) = \frac{3}{2}$

Step 3: Divide by $\frac{dx}{dt}$

$\frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}$

Answer: $\frac{d^2y}{dx^2} = \frac{3}{4t}$

Arc Length of Parametric Curves

The arc length from $t = a$ to $t = b$ is:

$L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$

Think: Speed = $\sqrt{(dx/dt)^2 + (dy/dt)^2}$, integrate over time!

Where This Comes From

Infinitesimal arc length: $ds = \sqrt{(dx)^2 + (dy)^2}$

$= \sqrt{\left(\frac{dx}{dt}dt\right)^2 + \left(\frac{dy}{dt}dt\right)^2}$

$= \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$

Integrate from $t = a$ to $t = b$!

Example 5: Arc Length of Circle

Find the circumference of $x = r\cos t$, $y = r\sin t$ for $0 \leq t \leq 2\pi$.

Step 1: Find derivatives

$\frac{dx}{dt} = -r\sin t, \quad \frac{dy}{dt} = r\cos t$

Step 2: Compute the integrand

$\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{r^2\sin^2 t + r^2\cos^2 t}$

$= \sqrt{r^2(\sin^2 t + \cos^2 t)} = \sqrt{r^2} = r$

Step 3: Integrate

$L = \int_0^{2\pi} r\,dt = r[t]_0^{2\pi} = r(2\pi - 0) = 2\pi r$

Answer: $2\pi r$ (the circumference formula!) ✓

Example 6: Arc Length with Integration

Find the arc length of $x = t^2$, $y = \frac{2}{3}t^3$ from $t = 0$ to $t = 1$.

Step 1: Find derivatives

$\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 2t^2$

Step 2: Set up integral

$L = \int_0^1 \sqrt{(2t)^2 + (2t^2)^2}\,dt$

$= \int_0^1 \sqrt{4t^2 + 4t^4}\,dt$

$= \int_0^1 \sqrt{4t^2(1 + t^2)}\,dt$

$= \int_0^1 2t\sqrt{1 + t^2}\,dt$

Step 3: Use substitution

Let $u = 1 + t^2$, $du = 2t\,dt$

When $t = 0$: $u = 1$ When $t = 1$: $u = 2$

$L = \int_1^2 \sqrt{u}\,du = \left[\frac{2u^{3/2}}{3}\right]_1^2$

$= \frac{2(2)^{3/2}}{3} - \frac{2(1)^{3/2}}{3} = \frac{2(2\sqrt{2})}{3} - \frac{2}{3}$

$= \frac{4\sqrt{2} - 2}{3}$

Answer: $\frac{4\sqrt{2} - 2}{3}$ or $\frac{2(2\sqrt{2} - 1)}{3}$

Surface Area of Revolution

When rotating a parametric curve around the x-axis from $t = a$ to $t = b$:

$S = 2\pi \int_a^b y\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$

Around the y-axis:

$S = 2\pi \int_a^b x\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$

⚠️ Common Mistakes

Mistake 1: Flipping the Fraction

WRONG: $\frac{dy}{dx} = \frac{dx/dt}{dy/dt}$

RIGHT: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

The derivative you want is on top!

Mistake 2: Forgetting to Check Conditions

For horizontal tangent: $\frac{dy}{dt} = 0$ AND $\frac{dx}{dt} \neq 0$

Don't forget to verify the second condition!

Mistake 3: Wrong Arc Length Formula

WRONG: $L = \int \sqrt{(dx)^2 + (dy)^2}$

RIGHT: $L = \int \sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$

Need to integrate with respect to the parameter!

Mistake 4: Second Derivative Error

$\frac{d^2y}{dx^2} \neq \frac{d^2y/dt^2}{d^2x/dt^2}$

Must use: $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt}$

Summary of Formulas

First Derivative:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Second Derivative:

$\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt}$

Arc Length:

$L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$

📝 Practice Strategy

1. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ first
2. For slope: Divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$
3. For horizontal tangent: Set $\frac{dy}{dt} = 0$, check $\frac{dx}{dt} \neq 0$
4. For vertical tangent: Set $\frac{dx}{dt} = 0$, check $\frac{dy}{dt} \neq 0$
5. For arc length: Use formula with square root, often needs u-substitution
6. Check your setup before integrating
7. Simplify under the square root if possible

$\frac{dx}{dt} = 3t^2 - 3$

$\frac{dy}{dt} = 2t$

$\frac{dy}{dx} = \frac{2t}{3t^2 - 3} = \frac{2t}{3(t^2 - 1)}$

Part (b): $\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right] = \frac{d/dt[dy/dx]}{dx/dt}$

$\frac{d}{dt}\left[\frac{2t}{3(t^2-1)}\right] = \frac{2 \cdot 3(t^2-1) - 2t \cdot 6t}{9(t^2-1)^2}$

$= \frac{6t^2 - 6 - 12t^2}{9(t^2-1)^2} = \frac{-6t^2 - 6}{9(t^2-1)^2} = \frac{-6(t^2+1)}{9(t^2-1)^2}$

$\frac{d^2y}{dx^2} = \frac{-6(t^2+1)/[9(t^2-1)^2]}{3(t^2-1)} = \frac{-6(t^2+1)}{27(t^2-1)^3} = \frac{-2(t^2+1)}{9(t^2-1)^3}$

Part (c): At $t = 2$:

$x = 8 - 6 = 2$, $y = 4$

$\frac{dy}{dx} = \frac{4}{3(3)} = \frac{4}{9}$

Tangent line: $y - 4 = \frac{4}{9}(x - 2)$

$y = \frac{4}{9}x - \frac{8}{9} + 4 = \frac{4}{9}x + \frac{28}{9}$

Step 2: For vertical tangent, set $\frac{dx}{dt} = 0$

$-\sin t = 0$ $\sin t = 0$

In $[0, 2\pi]$: $t = 0, \pi, 2\pi$

Step 3: Check that $\frac{dy}{dt} \neq 0$ at these values

At $t = 0$: $\frac{dy}{dt} = 2\cos(0) = 2 \neq 0$ ✓

At $t = \pi$: $\frac{dy}{dt} = 2\cos(2\pi) = 2 \neq 0$ ✓

At $t = 2\pi$: $\frac{dy}{dt} = 2\cos(4\pi) = 2 \neq 0$ ✓

Step 4: Find the points

At $t = 0$: $x = \cos 0 = 1$, $y = \sin 0 = 0$ → $(1, 0)$

At $t = \pi$: $x = \cos \pi = -1$, $y = \sin 2\pi = 0$ → $(-1, 0)$

At $t = 2\pi$: $x = \cos 2\pi = 1$, $y = \sin 4\pi = 0$ → $(1, 0)$

(Note: $t=0$ and $t=2\pi$ give the same point)

Answer: Vertical tangents at $(1, 0)$ and $(-1, 0)$.

$\frac{dy}{dt} = e^t\sin t + e^t\cos t = e^t(\sin t + \cos t)$

Step 2: Compute $(dx/dt)^2 + (dy/dt)^2$

$(dx/dt)^2 = e^{2t}(\cos t - \sin t)^2 = e^{2t}(\cos^2 t - 2\sin t\cos t + \sin^2 t)$

$= e^{2t}(1 - 2\sin t\cos t)$

$(dy/dt)^2 = e^{2t}(\sin t + \cos t)^2 = e^{2t}(\sin^2 t + 2\sin t\cos t + \cos^2 t)$

$= e^{2t}(1 + 2\sin t\cos t)$

Step 3: Add

$(dx/dt)^2 + (dy/dt)^2 = e^{2t}(1 - 2\sin t\cos t) + e^{2t}(1 + 2\sin t\cos t)$

$= e^{2t}(2) = 2e^{2t}$

Step 4: Set up arc length integral

$L = \int_0^{\pi/2} \sqrt{2e^{2t}}\,dt = \int_0^{\pi/2} e^t\sqrt{2}\,dt$

$= \sqrt{2}\int_0^{\pi/2} e^t\,dt$

Step 5: Integrate

$= \sqrt{2}[e^t]_0^{\pi/2}$

$= \sqrt{2}(e^{\pi/2} - e^0)$

$= \sqrt{2}(e^{\pi/2} - 1)$

Answer: $\sqrt{2}(e^{\pi/2} - 1)$