Tags: parametric equations, polar coordinates, polar curves, cardioid, rose curve, polar area, polar arc length, inverse function derivative, logarithmic differentiation, chain rule, log properties, Calculus II, Texas A&M, MATH 172, MATH 2414
This covers parametric curves (slope, eliminating the parameter), polar coordinates (identifying curves, computing area and arc length), and a few derivative techniques that come up repeatedly on Calc II exams: inverse function derivatives, logarithmic differentiation, and chain rule applications like d/dx[ln(cos(x))]. These topics often appear in the multiple-choice section, so fast recognition matters.
Parametric equations
A pair of equations x = f(t), y = g(t) that define a curve by tracing out points as the parameter t varies.
Parametric slope
dy/dx = (dy/dt)/(dx/dt) = g'(t)/f'(t). This is the slope of the tangent line to a parametric curve.
Polar coordinates
A coordinate system where each point is described by (r, θ): distance from the origin and angle from the positive x-axis.
Cardioid
A heart-shaped polar curve of the form r = a(1 + cos θ) or r = a(1 + sin θ).
Rose curve
A polar curve of the form r = a sin(nθ) or r = a cos(nθ). If n is odd, there are n petals. If n is even, there are 2n petals.
Improper integral
An integral with an infinite limit or an integrand with a discontinuity. Evaluated as a limit.
Inverse function derivative
If f is invertible and f(a) = b, then (f⁻¹)'(b) = 1/f'(a).
Logarithmic differentiation
A technique for differentiating functions of the form y = [f(x)]^g(x) by taking ln of both sides first.
Slope of a parametric curve:
dy/dx = g'(t)/f'(t)
This follows from the chain rule. It is not f'(t)/g'(t), which is a common exam trap.
Eliminating the parameter to identify curve type:
Use trig identities or algebraic manipulation to write the curve in Cartesian form.
Worked example: x = 5cos(t), y = 6sin(t)
cos(t) = x/5, sin(t) = y/6
cos²(t) + sin²(t) = 1 gives (x/5)² + (y/6)² = 1
This is an ellipse with semi-axes 5 and 6
If the coefficients were equal (e.g. x = 5cos(t), y = 5sin(t)), it would be a circle.
r = 2cos(θ)
Multiply both sides by r: r² = 2r cos(θ)
Convert: x² + y² = 2x
Complete the square: (x − 1)² + y² = 1
This is a circle of radius 1 centred at (1, 0)
r = a(1 + cos θ): cardioid
A symmetric, heart-shaped curve passing through the origin.
r = sin(2θ): rose curve
Since n = 2 (even), this rose has 2n = 4 petals. One full petal is traced for θ ∈ [0, π/2].
The area enclosed by a polar curve from θ = α to θ = β:
A = ∫ₐᵝ (1/2)r² dθ
Worked example: Area of one loop of r = sin(2θ)
One petal spans θ ∈ [0, π/2]
A = ∫₀^(π/2) (1/2) sin²(2θ) dθ
Use the identity sin²(u) = (1 − cos(2u))/2
A = (1/4) ∫₀^(π/2) (1 − cos(4θ)) dθ
= (1/4)[θ − sin(4θ)/4] from 0 to π/2
= (1/4)(π/2) = π/8
L = ∫ₐᵝ √(r² + (dr/dθ)²) dθ
Worked example: Length of r = 4(1 + cos θ) for θ ∈ [0, 2π]
r = 4 + 4cos θ, dr/dθ = −4sin θ
r² + (dr/dθ)² = 16(1 + cos θ)² + 16sin²θ = 16(2 + 2cos θ)
√(16(2 + 2cos θ)) = 4√(2 + 2cos θ)
Use the half-angle identity: 1 + cos θ = 2cos²(θ/2)
Integrand becomes 4 · 2|cos(θ/2)| = 8|cos(θ/2)|
By symmetry: L = 2 · 8∫₀^π cos(θ/2) dθ = 16[2sin(θ/2)] from 0 to π = 32
The half-angle identity is the key step that makes this tractable. Without it, the integral is unpleasant.
Evaluate by replacing the infinite limit with a variable and taking a limit.
Worked example: ∫₁^∞ 1/x² dx
= limₜ→∞ ∫₁ᵗ x⁻² dx = limₜ→∞ [−1/x] from 1 to t
= limₜ→∞ (−1/t + 1) = 1
The integral converges to 1
Quick reference: ∫₁^∞ 1/xᵖ dx converges when p > 1 and diverges when p ≤ 1 (same rule as the p-series).
If f(a) = b, then (f⁻¹)'(b) = 1/f'(a).
The steps are always the same:
Given b, solve f(a) = b for a
Compute f'(a)
The answer is 1/f'(a)
Worked example: f(x) = x³ + x, find (f⁻¹)'(2)
Solve x³ + x = 2. By inspection, x = 1
f'(x) = 3x² + 1, so f'(1) = 4
(f⁻¹)'(2) = 1/4
Used for expressions of the form y = [f(x)]^g(x), where both the base and exponent depend on x.
Worked example: Differentiate y = xˣ
Take ln: ln(y) = x ln(x)
Differentiate implicitly: (1/y) y' = ln(x) + x · (1/x) = ln(x) + 1
Multiply by y: y' = xˣ(ln(x) + 1)
Worked example: Find d/dx[ln(cos(x))]
d/dx[ln(cos(x))] = (1/cos(x)) · (−sin(x)) = −tan(x)
These are tested directly on exams and are used as steps inside other problems:
ln(a) − ln(b) = ln(a/b)
ln(a) + ln(b) = ln(ab)
ln(aⁿ) = n ln(a)
The trap: ln(a) − ln(b) ≠ ln(a − b). This is a common wrong answer on multiple choice.
Formula | Expression |
|---|---|
Parametric slope | dy/dx = (dy/dt)/(dx/dt) |
Polar area | A = ∫(1/2)r² dθ |
Polar arc length | L = ∫√(r² + (dr/dθ)²) dθ |
Inverse function derivative | (f⁻¹)'(b) = 1/f'(a) where f(a) = b |
Half-angle identity | 1 + cos θ = 2cos²(θ/2) |
Ellipse (parametric) | x = a cos(t), y = b sin(t) → (x/a)² + (y/b)² = 1 |
⚠️ Parametric slope is dy/dx = (dy/dt)/(dx/dt), not the other way round. This appears as a distractor on nearly every exam.
⚠️ For polar area, the formula has a factor of 1/2 in front: A = ∫(1/2)r² dθ. Forgetting the 1/2 doubles your answer.
⚠️ When finding the area of one petal of a rose curve, get the bounds right. For r = sin(2θ), one petal is θ ∈ [0, π/2], not [0, 2π].
⚠️ The polar curve r = 2cos(θ) is a circle centred at (1, 0), not at the origin. Multiply both sides by r and convert to Cartesian to see why.
⚠️ For improper integrals, always write the limit notation explicitly. Jumping straight to the antiderivative without the limit loses marks.
⚠️ ln(a − b) ≠ ln(a) − ln(b). The correct identity is ln(a/b) = ln(a) − ln(b).
Q: For x = f(t), y = g(t), what is dy/dx?
A: g'(t)/f'(t). It is (dy/dt) divided by (dx/dt).
Q: What type of curve is x = 5cos(t), y = 6sin(t)?
A: An ellipse. Eliminating the parameter gives (x/5)² + (y/6)² = 1.
Q: What curve does r = 2cos(θ) represent?
A: A circle of radius 1 centred at (1, 0).
Q: Compute the area of one loop of r = sin(2θ).
A: π/8. Integrate (1/2)sin²(2θ) from 0 to π/2 using the double-angle identity.
Q: Find the length of the cardioid r = 4(1 + cos θ).
A: 32. Use the half-angle identity to simplify, then integrate.
Q: Evaluate ∫₁^∞ 1/x² dx.
A: 1. The integral converges since p = 2 > 1.
Q: If f(x) = x³ + x, find (f⁻¹)'(2).
A: 1/4. f(1) = 2, f'(1) = 4, so the answer is 1/4.
Q: Differentiate y = xˣ.
A: y' = xˣ(ln(x) + 1). Use logarithmic differentiation.
Q: Find d/dx[ln(cos(x))].
A: −tan(x). Chain rule: (1/cos(x)) · (−sin(x)).
Q: What is ln(a) − ln(b)?
A: ln(a/b). Not ln(a − b).
parametric equations, parametric slope, parametric derivative, eliminating the parameter, ellipse parametric, polar coordinates, polar area, polar arc length, cardioid, rose curve, r = 2cos theta, half-angle identity, improper integrals, convergent integral, p-integral, inverse function derivative, logarithmic differentiation, x to the x, chain rule, log properties, ln rules, Calc II polar, MATH 2414, MATH 172