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🎯⭐ INTERACTIVE LESSON

Parametric Equations

Learn step-by-step with interactive practice!

Parametric Equations - Complete Interactive Lesson

Part 1: Parametric Basics

📈 Introduction to Parametric Equations

Part 1 of 7

What Are Parametric Equations?

Instead of y=f(x)y = f(x), we describe curves using a parameter tt:

x=f(t),y=g(t)x = f(t), \quad y = g(t)

As tt varies, the point (x,y)(x, y) traces a curve in the plane.

Why Use Parameters?

  • Direction & timing: parametric curves have a built-in direction (as tt increases)
  • Multiple yy values: can describe curves that fail the vertical line test (circles, loops)
  • Physical meaning: tt often represents time — the curve shows an object's path

Example: A Circle

x=cos⁡t,y=sin⁡t,0≤t≤2πx = \cos t, \quad y = \sin t, \quad 0 \leq t \leq 2\pi

Traces the unit circle counterclockwise starting at (1,0)(1, 0).

At t=0t = 0: (1,0)(1, 0). At t=π2t = \frac{\pi}{2}: (0,1)(0, 1). At t=πt = \pi: (−1,0)(-1, 0).

📝 Common Parametric Representations

Lines

x=x0+at,y=y0+btx = x_0 + at, \quad y = y_0 + bt Direction: (a,b)(a, b). Passes through (x0,y0)(x_0, y_0) at t=0t = 0.

Parabolas

x=t,y=t2x = t, \quad y = t^2 This is just y=x2y = x^2 parametrized with t=xt = x.

Ellipses

x=acos⁡t,y=bsin⁡t,0≤t≤2πx = a\cos t, \quad y = b\sin t, \quad 0 \leq t \leq 2\pi Semi-axes aa and bb. Since x2a2+y2b2=cos⁡2t+sin⁡2t=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = \cos^2 t + \sin^2 t = 1.

Eliminating the Parameter

To find the rectangular (Cartesian) equation:

  1. Solve one equation for tt
  2. Substitute into the other

Example: x=2t+1,  y=t−3x = 2t + 1, \; y = t - 3. From xx: t=x−12t = \frac{x-1}{2}. Then y=x−12−3=x−72y = \frac{x-1}{2} - 3 = \frac{x-7}{2} → a line.

🔄 Orientation and Domain

Direction Matters

The same curve can have different orientations:

  • x=t,  y=t2x = t, \; y = t^2 for t∈[0,2]t \in [0, 2] → traces left to right
  • x=2−t,  y=(2−t)2x = 2-t, \; y = (2-t)^2 for t∈[0,2]t \in [0, 2] → traces right to left

Same parabolic arc, opposite direction!

Restricting the Parameter

The domain of tt controls which portion of the curve is drawn:

  • x=cos⁡t,  y=sin⁡tx = \cos t, \; y = \sin t for 0≤t≤π0 \leq t \leq \pi → upper semicircle only
  • x=cos⁡t,  y=sin⁡tx = \cos t, \; y = \sin t for 0≤t≤4π0 \leq t \leq 4\pi → circle traced twice

💡 Key Insight: Different parametrizations can produce the same geometric curve but with different starting points, speeds, and directions.

Parametric Basics 🎯

Evaluate & Eliminate 🧮

1) x=2t+1,  y=3t−2x = 2t + 1, \; y = 3t - 2. At t=3t = 3: the xx-coordinate is?

2) x=cos⁡t,  y=sin⁡tx = \cos t, \; y = \sin t. At t=π3t = \frac{\pi}{3}: xx = ? (Enter as a fraction like "1/2")

3) Eliminate tt from x=t+1,  y=t2+2tx = t + 1, \; y = t^2 + 2t. Express yy in terms of xx: y=x2+y = x^2 + ?x+x + ? (Enter the constant term, like "-1")

Identify the Curve 🔽

Exit Quiz ✅

Part 2: Graphing Parametric Curves

📐 Slopes & Tangent Lines for Parametric Curves

Part 2 of 7

The Parametric Derivative

For x=f(t),  y=g(t)x = f(t), \; y = g(t), the slope of the tangent line is:

dydx=dy/dtdx/dt=g′(t)f′(t)(provided f′(t)≠0)\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)} \quad \text{(provided } f'(t) \neq 0\text{)}

Key Cases

ConditionGeometric Meaning
dydt=0,  dxdt≠0\frac{dy}{dt} = 0, \; \frac{dx}{dt} \neq 0Horizontal tangent
dxdt=0,  dydt≠0\frac{dx}{dt} = 0, \; \frac{dy}{dt} \neq 0Vertical tangent
Both zeroNeed further analysis (possible cusp)

📝 Example: Tangent Line to a Cycloid

A cycloid is traced by a point on a rolling circle:

x=t−sin⁡t,y=1−cos⁡tx = t - \sin t, \quad y = 1 - \cos t

Find the slope at t=π2t = \frac{\pi}{2}.

dxdt=1−cos⁡t=1−0=1\frac{dx}{dt} = 1 - \cos t = 1 - 0 = 1 at t=π2t = \frac{\pi}{2}

dydt=sin⁡t=1\frac{dy}{dt} = \sin t = 1 at t=π2t = \frac{\pi}{2}

dydx=11=1\frac{dy}{dx} = \frac{1}{1} = 1

Point: x=π2−1,  y=1x = \frac{\pi}{2} - 1, \; y = 1.

Tangent line: y−1=1(x−(π2−1))y - 1 = 1\left(x - (\frac{\pi}{2}-1)\right)

Where are horizontal tangents? sin⁡t=0  ⟹  t=nπ\sin t = 0 \implies t = n\pi (nn integer).

At t=πt = \pi: point is (π,2)(\pi, 2) — top of the arch, slope = 0. ✓

📊 Second Derivative & Concavity

The second derivative for parametric curves:

d2ydx2=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}

Steps:

  1. Find dydx=y′(t)x′(t)\frac{dy}{dx} = \frac{y'(t)}{x'(t)}
  2. Differentiate this with respect to tt: ddt(dydx)\frac{d}{dt}\left(\frac{dy}{dx}\right)
  3. Divide by dxdt\frac{dx}{dt}

Example: x=t2,  y=t3x = t^2, \; y = t^3

dydx=3t22t=3t2\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}

ddt(3t2)=32\frac{d}{dt}\left(\frac{3t}{2}\right) = \frac{3}{2}

d2ydx2=3/22t=34t\frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}

Concave up when t>0t > 0, concave down when t<0t < 0.

Derivative Quiz 🎯

Compute Slopes 🧮

1) x=3t,  y=t2−1x = 3t, \; y = t^2 - 1. What is dydx\frac{dy}{dx} at t=3t = 3? (Enter as a fraction or whole number)

2) x=et,  y=e−tx = e^t, \; y = e^{-t}. dydx\frac{dy}{dx} at t=0t = 0 = ?

3) x=t+sin⁡t,  y=1−cos⁡tx = t + \sin t, \; y = 1 - \cos t. dydx\frac{dy}{dx} at t=π2t = \frac{\pi}{2} = ? (Enter as a fraction)

Tangent Properties 🔽

Exit Quiz ✅

Part 3: Eliminating the Parameter

📏 Arc Length of Parametric Curves

Part 3 of 7

The Arc Length Formula

For a smooth curve x=f(t),  y=g(t)x = f(t), \; y = g(t) from t=at = a to t=bt = b:

L=∫ab(dxdt)2+(dydt)2 dtL = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt

Intuition

At each instant, the point moves by:

  • Δx≈dxdtΔt\Delta x \approx \frac{dx}{dt}\Delta t (horizontal)
  • Δy≈dydtΔt\Delta y \approx \frac{dy}{dt}\Delta t (vertical)

By the Pythagorean theorem: Δs≈(Δx)2+(Δy)2\Delta s \approx \sqrt{(\Delta x)^2 + (\Delta y)^2}

Summing up → the integral.

📝 Example 1: Circle Circumference

x=Rcos⁡t,  y=Rsin⁡t,  0≤t≤2πx = R\cos t, \; y = R\sin t, \; 0 \leq t \leq 2\pi

dxdt=−Rsin⁡t,dydt=Rcos⁡t\frac{dx}{dt} = -R\sin t, \quad \frac{dy}{dt} = R\cos t

L=∫02πR2sin⁡2t+R2cos⁡2t dt=∫02πR dt=2πR  ✓L = \int_0^{2\pi}\sqrt{R^2\sin^2 t + R^2\cos^2 t}\,dt = \int_0^{2\pi}R\,dt = 2\pi R \; ✓

Example 2: Line Segment

x=1+3t,  y=2+4t,  0≤t≤1x = 1 + 3t, \; y = 2 + 4t, \; 0 \leq t \leq 1

dxdt=3,dydt=4\frac{dx}{dt} = 3, \quad \frac{dy}{dt} = 4

L=∫019+16 dt=∫015 dt=5L = \int_0^1 \sqrt{9 + 16}\,dt = \int_0^1 5\,dt = 5

This matches the distance formula: 32+42=5\sqrt{3^2 + 4^2} = 5. ✓

Example 3: x=t2,  y=t3,  0≤t≤1x = t^2, \; y = t^3, \; 0 \leq t \leq 1

L=∫014t2+9t4 dt=∫01t4+9t2 dtL = \int_0^1\sqrt{4t^2 + 9t^4}\,dt = \int_0^1 t\sqrt{4+9t^2}\,dt

Let u=4+9t2u = 4+9t^2: du=18t dtdu = 18t\,dt

=118∫413u du=127[u3/2]413=1313−827= \frac{1}{18}\int_4^{13}\sqrt{u}\,du = \frac{1}{27}\left[u^{3/2}\right]_4^{13} = \frac{13\sqrt{13}-8}{27}

🚀 Speed Along a Parametric Curve

The speed at time tt is the rate of change of arc length:

speed=dsdt=(dxdt)2+(dydt)2\text{speed} = \frac{ds}{dt} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}

Example: Projectile Motion

x=v0t,  y=h0+v0t−12gt2x = v_0 t, \; y = h_0 + v_0 t - \frac{1}{2}gt^2

Speed: v02+(v0−gt)2\sqrt{v_0^2 + (v_0 - gt)^2}

At t=0t = 0: speed =v02+v02=v02= \sqrt{v_0^2 + v_0^2} = v_0\sqrt{2}

💡 Key: Speed is ALWAYS non-negative. It equals the magnitude of the velocity vector (dxdt,dydt)(\frac{dx}{dt}, \frac{dy}{dt}).

Arc Length Quiz 🎯

Arc Length Calculations 🧮

1) x=5t,  y=12tx = 5t, \; y = 12t from t=0t = 0 to t=3t = 3. Arc length = ?

2) The speed of a particle with x=3cos⁡t,  y=3sin⁡tx = 3\cos t, \; y = 3\sin t is constant at what value?

3) x=t,  y=23t3/2x = t, \; y = \frac{2}{3}t^{3/2} from t=0t = 0 to t=3t = 3. The integrand 1+t\sqrt{1+t} gives L=23[u3/2]14L = \frac{2}{3}[u^{3/2}]_1^4. Evaluate: LL = ? (Enter as a fraction like "14/3")

Arc Length Concepts 🔽

Exit Quiz ✅

Part 4: Parametric Motion

🎯 Projectile Motion & Applications

Part 4 of 7

Projectile Motion Equations

An object launched at angle α\alpha with initial speed v0v_0 from height h0h_0:

x=(v0cos⁡α) t,y=h0+(v0sin⁡α) t−12gt2x = (v_0\cos\alpha)\,t, \quad y = h_0 + (v_0\sin\alpha)\,t - \frac{1}{2}gt^2

where g≈9.8g \approx 9.8 m/s2m/s^{2} or 3232 ft/s2ft/s^{2}.

Key Quantities

QuantityFormula
Time of flightSolve y=0y = 0 (quadratic in tt)
Maximum heightAt t=v0sin⁡αgt = \frac{v_0\sin\alpha}{g} (when dydt=0\frac{dy}{dt} = 0)
Rangexx at landing time
Maximum rangeAt α=45°\alpha = 45° (on level ground)

📝 Example: Baseball Problem

A ball is hit at v0=128v_0 = 128 ft/s at angle α=30°\alpha = 30° from h0=3h_0 = 3 ft.

x=128cos⁡30° t=643 tx = 128\cos 30°\,t = 64\sqrt{3}\,t y=3+128sin⁡30° t−16t2=3+64t−16t2y = 3 + 128\sin 30°\,t - 16t^2 = 3 + 64t - 16t^2

Maximum height: dydt=64−32t=0  ⟹  t=2\frac{dy}{dt} = 64 - 32t = 0 \implies t = 2 s

ymax⁡=3+64(2)−16(4)=3+128−64=67y_{\max} = 3 + 64(2) - 16(4) = 3 + 128 - 64 = 67 ft

Time of flight: 3+64t−16t2=0  ⟹  16t2−64t−3=03 + 64t - 16t^2 = 0 \implies 16t^2 - 64t - 3 = 0

t=64+4096+19232=64+428832≈4.05t = \frac{64 + \sqrt{4096 + 192}}{32} = \frac{64 + \sqrt{4288}}{32} \approx 4.05 s

Range: x≈643(4.05)≈449x \approx 64\sqrt{3}(4.05) \approx 449 ft

⚾ That is a home run in most ballparks!

🔧 Other Applications of Parametric Equations

Particle on a Ferris Wheel

Center at (0,h)(0, h), radius RR, angular speed ω\omega:

x=Rcos⁡(ωt),y=h+Rsin⁡(ωt)x = R\cos(\omega t), \quad y = h + R\sin(\omega t)

Spirograph (Epitrochoid)

x=(R+r)cos⁡t−dcos⁡(R+rrt)x = (R+r)\cos t - d\cos\left(\frac{R+r}{r}t\right) y=(R+r)sin⁡t−dsin⁡(R+rrt)y = (R+r)\sin t - d\sin\left(\frac{R+r}{r}t\right)

Lissajous Figures

x=Asin⁡(at+δ),y=Bsin⁡(bt)x = A\sin(at + \delta), \quad y = B\sin(bt)

The shape depends on the frequency ratio a:ba:b and phase shift δ\delta.

  • a=b,δ=0a = b, \delta = 0: line
  • a=b,δ=π2a = b, \delta = \frac{\pi}{2}: ellipse
  • a=2,b=1a = 2, b = 1: figure-eight shapes

Applications Quiz 🎯

Projectile Calculations 🧮

A ball is launched at v0=80v_0 = 80 ft/s at α=45°\alpha = 45° from ground level (h0=0h_0 = 0, g=32g = 32 ft/s2ft/s^{2}).

1) The horizontal component of velocity v0cos⁡45°v_0\cos 45° = ? (Enter like "40sqrt2")

2) Time to reach max height = v0sin⁡αg\frac{v_0\sin\alpha}{g} = ? (Enter as a fraction like "5/2")

3) Maximum height = (v0sin⁡α)22g\frac{(v_0\sin\alpha)^2}{2g} = ? (whole number in ft)

Projectile Properties 🔽

Exit Quiz ✅

Part 5: Applications

🔄 Parametric Curves & Eliminating the Parameter

Part 5 of 7

Techniques for Eliminating the Parameter

Parametric FormStrategyRectangular Result
x=at+b,  y=ct+dx = at + b, \; y = ct + dSolve for tt from eitherLinear: y=mx+ky = mx + k
x=acos⁡t,  y=bsin⁡tx = a\cos t, \; y = b\sin tUse cos⁡2t+sin⁡2t=1\cos^2 t + \sin^2 t = 1x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
x=asec⁡t,  y=btan⁡tx = a\sec t, \; y = b\tan tUse sec⁡2t−tan⁡2t=1\sec^2 t - \tan^2 t = 1x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
x=tn,  y=f(t)x = t^n, \; y = f(t)t=x1/nt = x^{1/n}, substituteDepends on ff
x=et,  y=g(t)x = e^t, \; y = g(t)t=ln⁡xt = \ln x, substituteDepends on gg

📝 Worked Examples

Example 1: Trig Elimination

x=3+2cos⁡t,  y=−1+2sin⁡tx = 3 + 2\cos t, \; y = -1 + 2\sin t

(x−3)=2cos⁡t,  (y+1)=2sin⁡t(x-3) = 2\cos t, \; (y+1) = 2\sin t

(x−3)2+(y+1)2=4cos⁡2t+4sin⁡2t=4(x-3)^2 + (y+1)^2 = 4\cos^2 t + 4\sin^2 t = 4

Circle centered at (3,−1)(3, -1), radius 22.

Example 2: Exponential

x=e2t,  y=et+1x = e^{2t}, \; y = e^t + 1

Since x=e2t=(et)2x = e^{2t} = (e^t)^2 and et=y−1e^t = y - 1:

x=(y−1)2x = (y-1)^2, i.e., y=1+xy = 1 + \sqrt{x} (since et>0  ⟹  y>1e^t > 0 \implies y > 1)

Example 3: Watch the Domain!

x=t2,  y=tx = t^2, \; y = t → x=y2x = y^2 (full sideways parabola)

x=sin⁡2t,  y=sin⁡tx = \sin^2 t, \; y = \sin t → also x=y2x = y^2, but only −1≤y≤1-1 \leq y \leq 1 and 0≤x≤10 \leq x \leq 1

⚠️ The parametrization restricts which part of the Cartesian curve is actually traced!

✏️ Creating Parametric Equations

Given a Cartesian curve, find parametric equations. Multiple answers exist!

For y=x2−4x+3y = x^2 - 4x + 3:

  • Simple: x=t,  y=t2−4t+3x = t, \; y = t^2 - 4t + 3
  • Right to left: x=−t,  y=t2+4t+3x = -t, \; y = t^2 + 4t + 3
  • Shifted: x=t+2,  y=t2−1x = t + 2, \; y = t^2 - 1 (completing the square)

For a circle x2+y2=25x^2 + y^2 = 25:

  • Standard: x=5cos⁡t,  y=5sin⁡tx = 5\cos t, \; y = 5\sin t (CCW from (5,0)(5,0))
  • Clockwise: x=5cos⁡t,  y=−5sin⁡tx = 5\cos t, \; y = -5\sin t
  • Starting at top: x=5sin⁡t,  y=5cos⁡tx = 5\sin t, \; y = 5\cos t

For a line through (2,3)(2, 3) and (5,7)(5, 7):

Direction: (3,4)(3, 4). So x=2+3t,  y=3+4tx = 2 + 3t, \; y = 3 + 4t with t∈[0,1]t \in [0, 1] for the segment.

Elimination Quiz 🎯

Eliminate Parameters 🧮

1) x=2t−1,  y=4t+3x = 2t - 1, \; y = 4t + 3. Express yy in terms of xx: y=2x+y = 2x + ? (Enter the constant)

2) x=t3,  y=t2x = t^3, \; y = t^2. Express xx in terms of yy: x2=ynx^2 = y^n. What is nn? (whole number)

3) x=2cosh⁡t,  y=3sinh⁡tx = 2\cosh t, \; y = 3\sinh t. The curve is x24−y2n=1\frac{x^2}{4} - \frac{y^2}{n} = 1. What is nn?

Matching Curves 🔽

Exit Quiz ✅

Part 6: Problem-Solving Workshop

🌀 Special Parametric Curves

Part 6 of 7

Famous Curves with Parametric Equations

Cycloid — Point on rim of rolling circle (radius aa): x=a(t−sin⁡t),y=a(1−cos⁡t)x = a(t - \sin t), \quad y = a(1 - \cos t) Properties: arches from (0,0)(0,0) to (2πa,0)(2\pi a, 0), max height 2a2a.

Astroid — Point inside rolling circle: x=acos⁡3t,y=asin⁡3tx = a\cos^3 t, \quad y = a\sin^3 t Rectangular: x2/3+y2/3=a2/3x^{2/3} + y^{2/3} = a^{2/3}

Involute of a Circle — Unwinding string from circle: x=a(cos⁡t+tsin⁡t),y=a(sin⁡t−tcos⁡t)x = a(\cos t + t\sin t), \quad y = a(\sin t - t\cos t)

📝 The Cycloid in Depth

The cycloid x=a(t−sin⁡t),  y=a(1−cos⁡t)x = a(t - \sin t), \; y = a(1 - \cos t) has remarkable properties.

Slope

dydx=asin⁡ta(1−cos⁡t)=sin⁡t1−cos⁡t=cot⁡t2\frac{dy}{dx} = \frac{a\sin t}{a(1-\cos t)} = \frac{\sin t}{1-\cos t} = \cot\frac{t}{2}

At t=πt = \pi (top of arch): slope = cot⁡π2=0\cot\frac{\pi}{2} = 0 → horizontal tangent ✓

At t→0+t \to 0^+: slope →∞\to \infty → vertical tangent (cusp) ✓

Arc Length of One Arch

L=∫02πa2(1−cos⁡t)2+a2sin⁡2t dtL = \int_0^{2\pi}\sqrt{a^2(1-\cos t)^2 + a^2\sin^2 t}\,dt

Simplifies using 1−2cos⁡t+cos⁡2t+sin⁡2t=2(1−cos⁡t)=4sin⁡2t21-2\cos t + \cos^2 t + \sin^2 t = 2(1-\cos t) = 4\sin^2\frac{t}{2}:

L=∫02π2asin⁡t2 dt=[−4acos⁡t2]02π=−4a(−1−1)=8aL = \int_0^{2\pi}2a\sin\frac{t}{2}\,dt = \left[-4a\cos\frac{t}{2}\right]_0^{2\pi} = -4a(-1-1) = 8a

🎯 One arch of the cycloid has length exactly 8a8a — eight times the radius!

🎵 Lissajous Figures

x=Asin⁡(at+δ),y=Bsin⁡(bt)x = A\sin(at + \delta), \quad y = B\sin(bt)

The frequency ratio a:ba:b determines the shape:

Ratio a:ba:bShape
1:1,δ=01:1, \delta = 0Line segment (diagonal)
1:1,δ=π21:1, \delta = \frac{\pi}{2}Ellipse
1:21:2Figure-eight (or bowtie)
2:32:3Pretzel-like curve
3:43:4Complex knotted pattern

The number of lobes: up to aa lobes horizontally and bb lobes vertically.

Visualization Tip

Set A=B=1,δ=π2A = B = 1, \delta = \frac{\pi}{2}, and increment the ratio a:ba:b. The complexity increases — these patterns appear in oscilloscope traces and physics demonstrations.

Special Curves Quiz 🎯

Special Curve Calculations 🧮

1) Cycloid with a=3a = 3: arc length of one arch = 8a8a = ?

2) Cycloid with a=5a = 5: maximum height above baseline = 2a2a = ?

3) Astroid with a=8a = 8: at t=0t = 0, the point is at (x,y)=(?,0)(x, y) = (?, 0). What is xx?

Curve Identification 🔽

Exit Quiz ✅

Part 7: Review & Applications

🧩 Parametric Equations — Full Synthesis

Part 7 of 7

Complete Skill Set

TopicKey Idea
Parametrizationx=f(t),y=g(t)x = f(t), y = g(t); direction from increasing tt
Eliminating ttSolve for tt, use identities (sin⁡2+cos⁡2=1\sin^2+\cos^2=1, etc.)
Slopedydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}
Second derivatived2ydx2=(d/dt)(dy/dx)dx/dt\frac{d^2y}{dx^2} = \frac{(d/dt)(dy/dx)}{dx/dt}
Arc lengthL=∫(dx/dt)2+(dy/dt)2 dtL = \int\sqrt{(dx/dt)^2+(dy/dt)^2}\,dt
Speeddsdt=(dx/dt)2+(dy/dt)2\frac{ds}{dt} = \sqrt{(dx/dt)^2+(dy/dt)^2}
Projectilex=v0cos⁡α⋅t,  y=h0+v0sin⁡α⋅t−g2t2x = v_0\cos\alpha\cdot t, \; y = h_0 + v_0\sin\alpha\cdot t - \frac{g}{2}t^2
Special curvesCycloid, astroid, Lissajous, involute

🎓 Problem-Solving Flowchart

Given parametric equations, find...

Cartesian equation? → Eliminate tt (algebraic or trig identity)

  • Don't forget domain restrictions!

Slope at a point? → dy/dtdx/dt\frac{dy/dt}{dx/dt} at that tt value

Horizontal tangent? → dy/dt=0dy/dt = 0 (and dx/dt≠0dx/dt \neq 0)

Vertical tangent? → dx/dt=0dx/dt = 0 (and dy/dt≠0dy/dt \neq 0)

Arc length? → ∫(dx/dt)2+(dy/dt)2 dt\int\sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt

Concavity? → Compute d2ydx2\frac{d^2y}{dx^2} using the parametric formula

Common Mistakes

  • Forgetting that eliminating tt may lose domain information
  • Using d2ydx2=d2y/dt2d2x/dt2\frac{d^2y}{dx^2} = \frac{d^2y/dt^2}{d^2x/dt^2} (WRONG!)
  • Not checking dx/dt≠0dx/dt \neq 0 for horizontal tangents

Comprehensive Quiz 🎯

Mixed Calculations 🧮

1) x=t2,y=t3x = t^2, y = t^3. Find dydx\frac{dy}{dx} at t=2t = 2. (whole number)

2) x=5t,y=12tx = 5t, y = 12t from t=0t = 0 to t=4t = 4. Arc length = ?

3) Cycloid x=4(t−sin⁡t),y=4(1−cos⁡t)x = 4(t-\sin t), y = 4(1-\cos t). Arc length of one arch (8a8a) = ?

Final Concepts 🔽

Exit Quiz — Final ✅