Parametric equations express the coordinates \( x \) and \( y \) of a plane curve as separate functions of an independent variable called the parameter, typically denoted \( t \). Mastering parametric curves requires fluency in several interrelated skills: eliminating the parameter to recover a rectangular equation, analysing orientation and domain of the traced path, differentiating parametrically to find tangent and normal lines, computing concavity via second derivatives, and applying integral formulas to determine arc length and enclosed area. The problems below progress from foundational graphing tasks through mid-level calculus applications to challenging multi-step problems involving cycloids, surface area, and real-world modelling — covering every topic tested in Precalculus, Calculus I, and Calculus II courses.
Graphing Parametric Curves and Eliminating the Parameter
Before applying calculus, students must be able to convert a parametric representation into a recognisable rectangular form and accurately sketch the resulting curve with its correct orientation. This section builds that essential foundation, moving from parabolic curves through conics and trigonometric paths, with attention to domain, range, and direction of travel.
Problem 1: A Parabolic Curve with Restricted Parameter
Easy
A curve is defined by
\[ x = t^2 – 1, \qquad y = 2t + 3, \qquad -2 \le t \le 2. \]
- Eliminate \( t \) to find the rectangular equation, and state the domain and range.
- Sketch the curve and indicate its orientation.
Hint
View Solution
From \( y = 2t + 3 \): \( t = \dfrac{y-3}{2} \). Substituting into \( x = t^2 – 1 \):
\[ x = \left(\frac{y-3}{2}\right)^2 – 1 = \frac{(y-3)^2}{4} – 1. \]
When \( t = -2 \): \( x = 3,\, y = -1 \). When \( t = 2 \): \( x = 3,\, y = 7 \). When \( t = 0 \): \( x = -1,\, y = 3 \) (leftmost point). The domain is \( x \in [-1,\, 3] \) and the range is \( y \in [-1,\, 7] \).
Solution to question 2:
The curve is a horizontal parabola opening to the right with vertex at \((-1,\, 3)\). As \( t \) increases from \(-2\) to \( 2 \), the curve travels from \((3, -1)\) leftward to the vertex then rightward back to \((3, 7)\), traced bottom-to-top.
Problem 2: Trigonometric Parametrisation of an Ellipse
Medium
A curve is given by
\[ x = 3\cos\theta, \qquad y = 5\sin\theta, \qquad 0 \le \theta < 2\pi. \]
- Eliminate the parameter \( \theta \) and identify the conic section.
- State the semi-axes, describe the orientation, and determine how many times the curve is traced for \( 0 \le \theta < 2\pi \).
Hint
View Solution
From \( x = 3\cos\theta \) and \( y = 5\sin\theta \):
\[ \cos\theta = \frac{x}{3}, \qquad \sin\theta = \frac{y}{5}. \]
Applying \(\cos^2\theta + \sin^2\theta = 1\):
\[ \frac{x^2}{9} + \frac{y^2}{25} = 1. \]
This is an ellipse centred at the origin.
Solution to question 2:
The semi-axes are \( a = 3 \) (horizontal) and \( b = 5 \) (vertical). Starting at \((3, 0)\) when \(\theta = 0\), the curve moves counter-clockwise. It is traced exactly once for \( 0 \le \theta < 2\pi \).
Problem 3: Converting a Rectangular Equation to Parametric Form
Medium
The curve \( y = x^2 – 4x + 3 \) is defined for \( 0 \le x \le 4 \).
- Write two different sets of parametric equations that represent this curve.
- For each set, describe the orientation and the speed of traversal relative to the other.
Hint
View Solution
Set A (standard speed): Let \( t = x \), so \( 0 \le t \le 4 \).
\[ x = t, \qquad y = t^2 – 4t + 3. \]
Set B (half speed): Let \( x = 2t \), so \( 0 \le t \le 2 \).
\[ x = 2t, \qquad y = 4t^2 – 8t + 3. \]
Solution to question 2:
Both sets trace the same parabolic arc from \((0,\,3)\) to \((4,\,3)\) moving left to right. Set A traverses the path in one unit of \( t \) per unit of \( x \), while Set B takes twice as long — it is slower by a factor of two.
Differentiation: Tangent Lines and Normal Lines to Parametric Curves
The slope of a tangent line to a parametric curve is given by the chain-rule formula \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\), provided \(\dfrac{dx}{dt} \neq 0\). This section develops that formula through problems of increasing complexity, including horizontal and vertical tangents, normal lines, and cases where the curve crosses itself.
Problem 4: Finding a Tangent Line at a Given Parameter Value
Easy
Let the parametric curve be defined by
\[ x = t^2 + 1, \qquad y = t^3 – 3t. \]
- Find \(\dfrac{dy}{dx}\) in terms of \( t \).
- Find the equation of the tangent line at \( t = 2 \).
Hint
View Solution
\[ \frac{dx}{dt} = 2t, \qquad \frac{dy}{dt} = 3t^2 – 3. \]
\[ \frac{dy}{dx} = \frac{3t^2-3}{2t}, \quad t \neq 0. \]
Solution to question 2:
At \( t = 2 \): point \( (x,y) = (5,\, 2) \) and slope \( m = \dfrac{3(4)-3}{4} = \dfrac{9}{4} \). Using point-slope form:
\[ y – 2 = \frac{9}{4}(x – 5) \quad \Longrightarrow \quad y = \frac{9}{4}x – \frac{37}{4}. \]
Problem 5: Horizontal and Vertical Tangent Lines
Easy
For the parametric curve
\[ x = t^3 – 3t, \qquad y = t^2 – 1, \]
- Find all values of \( t \) at which the curve has a horizontal tangent line.
- Find all values of \( t \) at which the curve has a vertical tangent line.
Hint
View Solution
\[ \frac{dy}{dt} = 2t = 0 \;\Rightarrow\; t = 0. \]
Check: \( \dfrac{dx}{dt}\big|_{t=0} = -3 \neq 0 \). ✓ Horizontal tangent at \((0,\,-1)\).
Solution to question 2:
\[ \frac{dx}{dt} = 3t^2 – 3 = 0 \;\Rightarrow\; t = \pm 1. \]
Check \( dy/dt = 2t \neq 0 \) at both points. ✓ Vertical tangents at \( t = 1 \): point \((-2,\, 0)\), and at \( t = -1 \): point \((2,\, 0)\).
Problem 6: Tangent and Normal Lines on an Ellipse
Medium
An ellipse is parametrised by
\[ x = 4\cos\theta, \qquad y = 2\sin\theta, \qquad 0 \le \theta < 2\pi. \]
- Find the equations of the tangent line and the normal line to the curve at \(\theta = \dfrac{\pi}{4}\).
- Determine all \(\theta \in [0, 2\pi)\) where the tangent line is parallel to the line \( y = x \).
Hint
View Solution
\[ \frac{dx}{d\theta} = -4\sin\theta, \qquad \frac{dy}{d\theta} = 2\cos\theta. \]
\[ \frac{dy}{dx} = \frac{2\cos\theta}{-4\sin\theta} = -\frac{\cos\theta}{2\sin\theta}. \]
At \(\theta = \pi/4\): \( (x,y) = (2\sqrt{2},\, \sqrt{2}) \) and slope \( m_T = -\dfrac{1}{2} \).
Tangent line:
\[ y – \sqrt{2} = -\frac{1}{2}(x – 2\sqrt{2}) \;\Longrightarrow\; y = -\frac{x}{2} + 2\sqrt{2}. \]
Normal line (slope \( m_N = 2 \)):
\[ y – \sqrt{2} = 2(x – 2\sqrt{2}) \;\Longrightarrow\; y = 2x – 3\sqrt{2}. \]
Solution to question 2:
\[ -\frac{\cos\theta}{2\sin\theta} = 1 \;\Rightarrow\; \tan\theta = -\frac{1}{2}. \]
\[ \theta = \pi – \arctan\!\left(\tfrac{1}{2}\right) \approx 2.678 \quad \text{and} \quad \theta = 2\pi – \arctan\!\left(\tfrac{1}{2}\right) \approx 5.820. \]
Problem 7: Second Derivative and Concavity
Medium
The parametric curve is
\[ x = e^t, \qquad y = e^{2t} – 4e^t. \]
- Find \(\dfrac{d^2y}{dx^2}\) as a function of \( t \).
- Determine the intervals of \( t \) on which the curve is concave up and those on which it is concave down.
Hint
View Solution
\[ \frac{dx}{dt} = e^t, \qquad \frac{dy}{dt} = 2e^{2t} – 4e^t. \]
\[ \frac{dy}{dx} = \frac{2e^{2t}-4e^t}{e^t} = 2e^t – 4. \]
Differentiating with respect to \( t \): \(\dfrac{d}{dt}(2e^t – 4) = 2e^t\).
\[ \frac{d^2y}{dx^2} = \frac{2e^t}{e^t} = 2. \]
Solution to question 2:
Since \(\dfrac{d^2y}{dx^2} = 2 > 0\) for all \( t \), the curve is concave up everywhere. This is consistent with the rectangular form \( y = x^2 – 4x \), an upward-opening parabola.
Problem 8: Tangent Lines at a Self-Intersection
Hard
A curve is defined by
\[ x = t^2 – t, \qquad y = t^3 – 3t^2 + 2t. \]
- Find the self-intersection point of the curve by determining the two distinct values of \( t \) that produce the same \((x,y)\) point.
- Find the equations of the two tangent lines at the self-intersection point.
Hint
View Solution
From \( x(t_1) = x(t_2) \): \( t_1^2 – t_1 = t_2^2 – t_2 \), so \( (t_1-t_2)(t_1+t_2-1)=0 \). Since \( t_1 \neq t_2 \), we need \( t_1 + t_2 = 1 \).
Using \( t_2 = 1-t_1 \) and equating the \( y \)-values yields \( t_1 = 0 \) or \( t_1 = 1 \). The self-intersection point is:
\[ (x,y) = (0,\,0). \]
Solution to question 2:
\[ \frac{dy}{dx} = \frac{3t^2 – 6t + 2}{2t – 1}. \]
At \( t = 0 \): slope \( m_1 = \dfrac{2}{-1} = -2 \). Tangent: \( y = -2x \).
At \( t = 1 \): slope \( m_2 = \dfrac{3-6+2}{1} = -1 \). Tangent: \( y = -x \).
Area Under and Between Parametric Curves
The area enclosed by a parametric curve and the \(x\)-axis is computed via the formula \(A = \displaystyle\int_{t_1}^{t_2} y(t)\, x'(t)\, dt\), with careful attention to the direction of traversal and the sign of the result. The problems in this section move from enclosed ellipse area to the classical cycloid arch.
Problem 9: Area Enclosed by an Ellipse
Medium
The ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) is parametrised by \( x = a\cos t,\; y = b\sin t \) for \( 0 \le t \le 2\pi \).
- Use the parametric area formula to show that the area enclosed by the ellipse is \( \pi ab \).
- Apply the result to find the area enclosed by \( 4x^2 + 9y^2 = 36 \).
Hint
View Solution
With \( x = a\cos t \), \( x'(t) = -a\sin t \), and \( y = b\sin t \):
\[ A = \left|\int_0^{2\pi} b\sin t \cdot (-a\sin t)\, dt\right| = ab\int_0^{2\pi}\sin^2 t\, dt = ab \cdot \pi = \pi ab. \]
Solution to question 2:
Rewrite \( 4x^2 + 9y^2 = 36 \) as \(\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1\), so \( a = 3,\; b = 2 \). Area \( = \pi(3)(2) = 6\pi \).
Problem 10: Area of a Region Bounded by a Cycloid
Hard
A cycloid is generated by a circle of radius \( r = 1 \) rolling along the \(x\)-axis, giving the parametric equations
\[ x = t – \sin t, \qquad y = 1 – \cos t, \qquad 0 \le t \le 2\pi. \]
- Compute the area under one complete arch of the cycloid (between \( t = 0 \) and \( t = 2\pi \)).
- Express the result as a multiple of the area of the generating circle and interpret the geometric meaning.
Hint
View Solution
\[ x'(t) = 1 – \cos t, \qquad y(t) = 1 – \cos t. \]
\[ A = \int_0^{2\pi}(1-\cos t)^2\, dt = \int_0^{2\pi}\!\left(1 – 2\cos t + \cos^2 t\right)dt. \]
\[ = \left[t – 2\sin t + \frac{t}{2} + \frac{\sin 2t}{4}\right]_0^{2\pi} = 2\pi + \pi = 3\pi. \]
Solution to question 2:
The area of the generating circle is \( \pi r^2 = \pi \). The area under one arch equals \( 3\pi \), exactly three times the area of the rolling circle — a classical result due to Evangelista Torricelli (1644).
Arc Length of Parametric Curves
The arc length of a smooth parametric curve \( x = f(t),\; y = g(t) \) on \([\alpha, \beta]\) is given by
\( L = \displaystyle\int_{\alpha}^{\beta}\!\sqrt{\left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2}\, dt \).
These problems develop skill in setting up and evaluating that integral, including cases requiring trigonometric simplification and \( u \)-substitution.
Problem 11: Arc Length of a Parametric Semicircle
Medium
A semicircle of radius \( r = 5 \) is parametrised by
\[ x = 5\cos t, \qquad y = 5\sin t, \qquad 0 \le t \le \pi. \]
- Use the arc length formula to compute the length of the semicircle.
- State the general formula for the arc length of a full circle of radius \( r \) using this parametrisation.
Hint
View Solution
\[ \frac{dx}{dt} = -5\sin t, \qquad \frac{dy}{dt} = 5\cos t. \]
\[ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 25\sin^2 t + 25\cos^2 t = 25. \]
\[ L = \int_0^{\pi}\sqrt{25}\, dt = 5\pi. \]
Solution to question 2:
For a full circle (\( 0 \le t \le 2\pi \)): \( L = 5 \cdot 2\pi = 10\pi \). In general, for radius \( r \): \( L = 2\pi r \).
Problem 12: Arc Length of a Power Curve
Medium
Find the arc length of the curve defined by
\[ x = t^2, \qquad y = \frac{2t^3}{3}, \qquad 0 \le t \le \sqrt{3}. \]
- Set up the arc length integral.
- Evaluate the integral by an appropriate substitution.
Hint
View Solution
\[ \frac{dx}{dt} = 2t, \qquad \frac{dy}{dt} = 2t^2. \]
\[ L = \int_0^{\sqrt{3}}\sqrt{4t^2 + 4t^4}\, dt = \int_0^{\sqrt{3}} 2t\sqrt{1+t^2}\, dt. \]
Solution to question 2:
Let \( u = 1 + t^2 \), so \( du = 2t\, dt \). When \( t = 0 \): \( u = 1 \); when \( t = \sqrt{3} \): \( u = 4 \).
\[ L = \int_1^4 \sqrt{u}\, du = \left[\frac{2}{3}u^{3/2}\right]_1^4 = \frac{2}{3}(8 – 1) = \frac{14}{3}. \]
Problem 13: Arc Length of One Arch of a Cycloid
Hard
For the cycloid \( x = r(t – \sin t),\; y = r(1 – \cos t) \), \( 0 \le t \le 2\pi \), with \( r > 0 \):
- Set up the arc length integral and simplify the integrand using a half-angle identity.
- Evaluate the integral to show that the arc length of one complete arch is \( 8r \).
Hint
View Solution
\[ \frac{dx}{dt} = r(1-\cos t), \qquad \frac{dy}{dt} = r\sin t. \]
\[ \left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2 = r^2(1-\cos t)^2 + r^2\sin^2 t = r^2(2 – 2\cos t). \]
Using \( 1 – \cos t = 2\sin^2(t/2) \):
\[ \sqrt{2r^2(1-\cos t)} = 2r\sin\!\left(\frac{t}{2}\right), \quad 0 \le t \le 2\pi. \]
Solution to question 2:
\[ L = \int_0^{2\pi} 2r\sin\!\left(\frac{t}{2}\right)dt. \]
Let \( u = t/2 \), \( du = dt/2 \):
\[ L = 4r\int_0^{\pi}\sin u\, du = 4r\big[-\cos u\big]_0^{\pi} = 4r(1+1) = 8r. \]
Real-World Applications of Parametric Equations
Parametric equations are the natural language of motion: they encode position, speed, and direction simultaneously. This section models projectile motion and particle velocity along curved paths, applying both algebraic and calculus tools to interpret physical scenarios — topics heavily represented in AP Calculus BC and university Calculus II exams.
Problem 14: Projectile Motion
Easy
A ball is launched from ground level with initial speed \( v_0 = 40 \) m/s at an angle of \(30°\) above the horizontal. Neglecting air resistance, the position is given by
\[ x = v_0\cos(30°)\, t, \qquad y = v_0\sin(30°)\, t – \frac{1}{2}(9.8)t^2. \]
- Find the time \( t \) at which the ball reaches maximum height, and compute that maximum height.
- Find the horizontal range (the value of \( x \) when the ball returns to ground level).
Hint
View Solution
\[ \frac{dy}{dt} = 40\sin(30°) – 9.8t = 20 – 9.8t = 0 \;\Rightarrow\; t^* = \frac{20}{9.8} \approx 2.04\text{ s}. \]
\[ y_{\max} = 20(2.04) – 4.9(2.04)^2 \approx 40.8 – 20.4 = 20.4\text{ m}. \]
Solution to question 2:
Set \( y = 0 \): \( t(20 – 4.9t) = 0 \Rightarrow t = 0 \) or \( t = 20/4.9 \approx 4.08 \) s.
\[ x_{\text{range}} = 40\cos(30°)\cdot 4.08 = 40\cdot\frac{\sqrt{3}}{2}\cdot 4.08 \approx 141.4\text{ m}. \]
Problem 15: Speed Along a Parametric Path
Medium
A particle moves along the curve
\[ x(t) = \sin(2t), \qquad y(t) = \cos t, \qquad 0 \le t \le \pi. \]
- Find the speed of the particle (the magnitude of the velocity vector) as a function of \( t \).
- Determine the time(s) in \([0, \pi]\) at which the speed is a minimum, and state the minimum speed.
Hint
View Solution
\[ \frac{dx}{dt} = 2\cos(2t), \qquad \frac{dy}{dt} = -\sin t. \]
\[ v(t) = \sqrt{4\cos^2(2t) + \sin^2 t}. \]
Using \(\cos(2t) = 1 – 2\sin^2 t\):
\[ v(t) = \sqrt{4(1-2\sin^2 t)^2 + \sin^2 t}. \]
Solution to question 2:
Let \( u = \sin^2 t \in [0,1] \). Then \( v^2 = 16u^2 – 15u + 4 \). Setting the derivative equal to zero: \( 32u – 15 = 0 \Rightarrow u = \tfrac{15}{32} \). Since \( u = \sin^2 t \), we get \( t \approx 0.846 \) or \( t \approx \pi – 0.846 \). The minimum speed is:
\[ v_{\min} = \sqrt{16\!\left(\tfrac{15}{32}\right)^2 – 15\!\left(\tfrac{15}{32}\right) + 4} = \sqrt{4 – \tfrac{225}{128}} \approx 1.65\text{ units/s}. \]