Difficulty: Intermediate | Prerequisites: Calculus I and II (MATH 124/125), comfort with derivatives, integrals, and basic parametric equations.
Calculus III (MATH 126) extends everything you learnt about single-variable functions into the world of vector-valued functions, curves in space, and multivariable calculus. This quiz covers the foundational toolkit: how to find where vector functions are defined, how to integrate them component by component, and how to write equations for tangent lines to space curves. If you are comfortable differentiating and integrating ordinary functions from Calc I/II, you already have most of the machinery. The new layer is applying it to each component of a vector separately, then reassembling the results.
A vector function's domain is the intersection of the domains of its individual components. You integrate and differentiate vector functions component by component. Tangent lines to parametric curves use the derivative vector evaluated at a point as the direction, combined with the point itself.
Vector-valued function (vector function)
A function r(t) whose output is a vector rather than a scalar. Typically written as r(t) = f(t) i + g(t) j + h(t) k, where f, g, h are ordinary real-valued functions called the component functions.
In simple terms, instead of spitting out a single number, this function gives you a point (or direction) in 2D or 3D space for each value of t.
Domain of a vector function
The set of all values of the parameter t for which every component function is defined. You find each component's domain separately, then take the intersection.
Think of it as: the function only works where all its parts work simultaneously.
Parametric equations
A way of describing a curve by expressing each coordinate (x, y, z) as a separate function of a parameter t. For example, x = t², y = sin(t), z = e^t.
In simple terms, instead of y = f(x), you let a third variable t drive both x and y (and z) independently.
Tangent vector
The derivative r'(t) of a vector function, found by differentiating each component. At a given point, this vector points in the direction the curve is heading.
Think of it as the velocity of a particle travelling along the curve.
Indefinite integral of a vector function
The antiderivative taken component by component, plus a constant vector C = C₁ i + C₂ j + C₃ k.
In simple terms, you integrate each piece on its own and tack on a vector of arbitrary constants.
Definite integral of a vector function
Evaluate the definite integral of each component over the given interval, then assemble the results into a vector. No constant of integration needed.
The domain of r(t) = ⟨f(t), g(t), h(t)⟩ is dom(f) ∩ dom(g) ∩ dom(h). Work through each component, list its restrictions, then intersect.
Worked Example 1 (Quiz Q1)
r(t) = ⟨√(6 − t), (e^t − 8)/t, ln(t + 1)⟩
Component 1: √(6 − t) requires 6 − t ≥ 0, so t ≤ 6.
Component 2: (e^t − 8)/t requires t ≠ 0.
Component 3: ln(t + 1) requires t + 1 > 0, so t > −1.
Intersection: t > −1, t ≠ 0, and t ≤ 6.
Domain: (−1, 0) ∪ (0, 6].
Worked Example 2 (Quiz Q2)
r(t) = 7t i + 1/(t − 9) j
Component 1: 7t is defined for all real t.
Component 2: 1/(t − 9) requires t ≠ 9.
Domain: (−∞, 9) ∪ (9, ∞). That is answer (a).
Method summary
Square roots: set the radicand ≥ 0.
Logarithms: set the argument > 0.
Fractions: set the denominator ≠ 0.
Exponentials and polynomials: no restrictions.
Intersect all component domains at the end.
Integrate each component independently. For indefinite integrals, add a constant vector C. For definite integrals, evaluate each component's antiderivative at the bounds and subtract.
Worked Example 3 (Quiz Q5) – Indefinite integral
∫ (e^(9t) i + 8t j + ln(t) k) dt
i-component: ∫ e^(9t) dt = (1/9)e^(9t) + C₁
j-component: ∫ 8t dt = 4t² + C₂
k-component: ∫ ln(t) dt = t ln(t) − t + C₃ (integration by parts: u = ln(t), dv = dt)
Result: ((1/9)e^(9t) + C₁) i + (4t² + C₂) j + (t ln(t) − t + C₃) k
Often written with a single vector constant C.
Worked Example 4 (Quiz Q6) – Definite integral
r(t) = 7 i + 3t cos(πt) j + 4 sin(πt) k. Evaluate ∫₀¹ r(t) dt.
i-component: ∫₀¹ 7 dt = 7.
j-component: ∫₀¹ 3t cos(πt) dt. Use integration by parts. Let u = 3t, dv = cos(πt) dt. Then du = 3 dt, v = (1/π) sin(πt). Applying: [3t · (1/π) sin(πt)]₀¹ − ∫₀¹ (3/π) sin(πt) dt = 0 − (3/π)[−(1/π) cos(πt)]₀¹ = −(3/π²)(cos(π) − cos(0)) = −(3/π²)(−1 − 1) = 6/π².
k-component: ∫₀¹ 4 sin(πt) dt = 4[−(1/π) cos(πt)]₀¹ = (−4/π)(cos(π) − cos(0)) = (−4/π)(−1 − 1) = 8/π.
Result: 7 i + (6/π²) j + (8/π) k.
Key reminders
The integral of ln(t) requires integration by parts. This is a favourite exam question.
For definite integrals involving trig functions, watch the argument carefully. cos(πt) is not the same as cos(t).
Always double-check signs when evaluating cos at the bounds.
Given a curve x = f(t), y = g(t), z = h(t), the tangent line at a point uses:
The point itself: (f(t₀), g(t₀), h(t₀)).
The direction vector: (f'(t₀), g'(t₀), h'(t₀)).
Parametric equations of the tangent line: x = f(t₀) + f'(t₀)·s, y = g(t₀) + g'(t₀)·s, z = h(t₀) + h'(t₀)·s, where s is the new parameter.
Worked Example 5 (Quiz Q7)
x = t¹¹, y = t⁵, z = t⁶ at the point (1, 1, 1).
Step 1: Find the parameter value. We need t¹¹ = 1, t⁵ = 1, t⁶ = 1. All give t = 1.
Step 2: Compute the derivatives at t = 1.
x' = 11t¹⁰, so x'(1) = 11.
y' = 5t⁴, so y'(1) = 5.
z' = 6t⁵, so z'(1) = 6.
Step 3: Write the tangent line.
x = 1 + 11t
y = 1 + 5t
z = 1 + 6t
That is answer (c).
Common pitfalls
Forgetting to find the correct t₀ from the given point. If the point does not correspond to a real value of t, something has gone wrong.
Using t as both the curve parameter and the tangent-line parameter. The quiz reuses t, but conceptually they are different variables.
Sign errors when the derivative is negative at the point. The tangent line would have a minus sign in that component.
Domain of r(t) = ⟨f(t), g(t), h(t)⟩
dom(r) = dom(f) ∩ dom(g) ∩ dom(h)
Indefinite integral
∫ r(t) dt = (∫ f(t) dt) i + (∫ g(t) dt) j + (∫ h(t) dt) k + C
Definite integral
∫ₐᵇ r(t) dt = (∫ₐᵇ f(t) dt) i + (∫ₐᵇ g(t) dt) j + (∫ₐᵇ h(t) dt) k
Tangent line at t = t₀
Point: r(t₀). Direction: r'(t₀). Parametric form: r(t₀) + s · r'(t₀).
Integration by parts (for reference)
∫ u dv = uv − ∫ v du
Useful antiderivatives
∫ e^(at) dt = (1/a)e^(at) + C
∫ ln(t) dt = t ln(t) − t + C
∫ sin(at) dt = −(1/a)cos(at) + C
∫ cos(at) dt = (1/a)sin(at) + C
Students often take the union of component domains instead of the intersection. The vector function is only defined where all components work, so you intersect.
When integrating vector functions, some students add a single scalar constant C instead of a vector constant C = C₁ i + C₂ j + C₃ k. Each component gets its own constant.
Students sometimes forget that the domain restriction from a square root is ≥ 0 (not just > 0), while the restriction from a logarithm is strictly > 0. This distinction matters at boundary points.
A common error on tangent-line problems is plugging the given point coordinates directly into the derivative formula instead of first solving for the correct parameter value t₀.
⚠️ Domain questions appear on nearly every early Calc III quiz. They test whether you can handle each component's restrictions and correctly intersect them.
⚠️ Integration by parts for vector integrals is a favourite. Expect at least one component that requires it, particularly ln(t) or t·trig(t).
⚠️ Tangent-line questions test the full chain: find t₀, differentiate, write the parametric equations. Partial credit often depends on showing each step.
⚠️ On multiple-choice domain problems, wrong answers typically swap the excluded point or use the wrong inequality direction. Read each option against your computed domain carefully.
True or False: The domain of r(t) = ⟨t², ln(t)⟩ is all real numbers. (False: ln(t) requires t > 0.)
Fill in the blank: To integrate a vector function, you integrate each ______ separately. (component)
True or False: The tangent vector to a parametric curve is found by integrating r(t). (False: it is found by differentiating r(t).)
Fill in the blank: The domain of r(t) = ⟨1/t, √t⟩ is ______. (t > 0, i.e. (0, ∞))
True or False: An indefinite vector integral needs only one scalar constant of integration. (False: it needs a vector constant, one per component.)
Q: Find the domain of r(t) = ⟨√(t − 2), 1/(t − 5), e^t⟩.
A: √(t − 2) needs t ≥ 2. 1/(t − 5) needs t ≠ 5. e^t has no restriction. Intersection: [2, 5) ∪ (5, ∞).
Q: Evaluate ∫ (3e^(2t) i + cos(t) j) dt.
A: (3/2)e^(2t) i + sin(t) j + C.
Q: A curve is given by x = t³, y = t², z = t at the point (8, 4, 2). Find parametric equations for the tangent line.
A: At (8, 4, 2), t = 2. Derivatives: x' = 3t² = 12, y' = 2t = 4, z' = 1. Tangent line: x = 8 + 12s, y = 4 + 4s, z = 2 + s.
Q: Find the domain of r(t) = ⟨ln(t), t/(t² − 4)⟩.
A: ln(t) needs t > 0. t/(t² − 4) needs t ≠ ±2, but since t > 0 we only exclude t = 2. Domain: (0, 2) ∪ (2, ∞).
Q: Evaluate ∫₀² (t i + t² j + k) dt.
A: i-component: [t²/2]₀² = 2. j-component: [t³/3]₀² = 8/3. k-component: [t]₀² = 2. Result: 2 i + (8/3) j + 2 k.
Domains of vector functions connect directly to arc length and curvature calculations later in MATH 126, because those formulas require the derivatives to exist on the interval you are integrating over.
Integrating vector functions is the basis for finding position from velocity and velocity from acceleration in the particle-motion problems that appear in the next unit.
Tangent lines to parametric curves lead into the tangent and normal vectors (T, N, B), which form the Frenet-Serret frame used throughout the rest of the course.
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