Source: Midterm I Practice Exam
Difficulty: Intermediate | Prerequisites: Derivatives of single-variable functions, vectors, dot and cross products.
Vector-valued functions describe curves in 2D or 3D by giving a position vector for each value of a parameter t. The tangent line at a point tells you the instantaneous direction of travel, and curvature measures how sharply the curve bends at each point. These ideas extend single-variable differentiation into space and set the stage for motion problems (velocity, acceleration, normal and tangential components) that come later in Math 126.
Differentiate a vector function component by component to get the tangent vector. The tangent line at a point uses that vector as its direction. Curvature κ = |r'(t) × r''(t)| / |r'(t)|³ measures bending. To maximise curvature, treat κ(t) as a single-variable function and find where its derivative is zero.
Vector-valued function
A function r(t) = ⟨x(t), y(t), z(t)⟩ that assigns a position vector to each parameter value t. The output is a curve in space.
In simple terms, this means: a recipe that traces out a path as t changes.
Tangent vector
r'(t) = ⟨x'(t), y'(t), z'(t)⟩. The derivative of the position function, computed component by component. It points in the direction of motion along the curve.
Think of it as: the velocity vector if t represents time.
Tangent line
The line through a point on the curve in the direction of the tangent vector. Given as L(s) = r(t₀) + s · r'(t₀).
In simple terms, this means: the straight line that best approximates the curve at that point.
Curvature (κ)
A scalar measuring how fast the curve changes direction per unit arc length. Computed as κ(t) = |r'(t) × r''(t)| / |r'(t)|³.
Think of it as: a large κ means the curve bends sharply (like a tight turn); a small κ means it is nearly straight.
Given r(t) = ⟨2t + 1, t - 5, t² - t + 2⟩ and the point (7, -2, 8).
Step 1: Find t. Match each component: 2t + 1 = 7 gives t = 3. Check: t - 5 = 3 - 5 = -2 ✓. t² - t + 2 = 9 - 3 + 2 = 8 ✓. So t = 3.
Step 2: Compute r'(t). r'(t) = ⟨2, 1, 2t - 1⟩. At t = 3: r'(3) = ⟨2, 1, 5⟩.
Step 3: Write the tangent line. L(s) = ⟨7, -2, 8⟩ + s⟨2, 1, 5⟩, or equivalently: x = 7 + 2s, y = -2 + s, z = 8 + 5s.
Using the formula κ(t) = |r'(t) × r''(t)| / |r'(t)|³.
r'(t) = ⟨2, 1, 2t - 1⟩.
r''(t) = ⟨0, 0, 2⟩.
r'(t) × r''(t) = ⟨(1)(2) - (2t - 1)(0), (2t - 1)(0) - (2)(2), (2)(0) - (1)(0)⟩ = ⟨2, -4, 0⟩.
|r'(t) × r''(t)| = √(4 + 16 + 0) = √20 = 2√5.
|r'(t)| = √(4 + 1 + (2t - 1)²) = √(5 + (2t - 1)²).
κ(t) = 2√5 / (5 + (2t - 1)²)^(3/2).
Notice that the numerator is constant. This simplifies the maximisation step considerably.
Since κ(t) = 2√5 / (5 + (2t - 1)²)^(3/2), the numerator is fixed. So κ is maximised when the denominator is minimised.
The denominator (5 + (2t - 1)²)^(3/2) is smallest when (2t - 1)² = 0, i.e. t = 1/2.
Maximum curvature: κ(1/2) = 2√5 / 5^(3/2) = 2√5 / (5√5) = 2/5.
The point: r(1/2) = ⟨2(1/2) + 1, 1/2 - 5, 1/4 - 1/2 + 2⟩ = ⟨2, -9/2, 7/4⟩.
\mathbf{r}'(t) = \langle x'(t),\; y'(t),\; z'(t) \rangle\text{Tangent line: } \mathbf{L}(s) = \mathbf{r}(t_0) + s\,\mathbf{r}'(t_0)\kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}Curvature determines how sharply a road or railway track bends, which directly affects safe speed limits. In robotics and CNC machining, the tangent vector at each point of a tool path dictates the direction of motion, and curvature limits how fast the tool can move without overshooting. In physics, curvature of a particle's trajectory is related to the centripetal acceleration it experiences.
Students often confuse the parameter t with arc length s. The tangent line uses a new parameter (s here) for the line itself, not the same t from the curve.
When finding the value of t corresponding to a given point, students sometimes solve only one component and skip verification. You must check all three components match at the same t.
The curvature formula uses the cross product of r' and r'', not the dot product. Using the dot product gives a scalar that has nothing to do with curvature.
To maximise κ, students sometimes take the derivative of the full expression and set it to zero. When the numerator is constant (as in this exam problem), it is much simpler to minimise the denominator instead.
⚠️ The tangent-line problem is worth 7 points on this exam. Finding the correct t by matching the given point to r(t) is where most errors happen.
⚠️ In the curvature computation, the cross product r' × r'' often simplifies nicely. Do not skip simplifying before plugging into the magnitude.
⚠️ When asked for maximum curvature, state both the maximum value of κ and the point where it occurs. Missing either part costs marks.
⚠️ All answers must be exact. Leave curvature as 2/5, not 0.4.
r(t) = ⟨t², t, 1⟩. What is r'(t)? (⟨2t, 1, 0⟩.)
True or false: curvature can be negative. (False. It is always non-negative.)
The curvature of a straight line is ___. (Zero.)
If r''(t) = ⟨0, 0, 0⟩ for all t, the curve is a ___. (Straight line.)
True or false: to write the tangent line, you need both a point and a direction vector. (True.)
Q: Given r(t) = ⟨cos t, sin t, t⟩, find the tangent line at t = π/2.
A: r(π/2) = ⟨0, 1, π/2⟩. r'(t) = ⟨-sin t, cos t, 1⟩. r'(π/2) = ⟨-1, 0, 1⟩. Tangent line: L(s) = ⟨0, 1, π/2⟩ + s⟨-1, 0, 1⟩.
Q: Compute the curvature of r(t) = ⟨t, t², 0⟩.
A: r'(t) = ⟨1, 2t, 0⟩, r''(t) = ⟨0, 2, 0⟩. r' × r'' = ⟨0, 0, 2⟩. |r' × r''| = 2. |r'| = √(1 + 4t²). κ(t) = 2/(1 + 4t²)^(3/2).
Q: For the curve in the previous question, where does maximum curvature occur?
A: Minimise the denominator (1 + 4t²)^(3/2). This is smallest when t = 0. κ(0) = 2/1 = 2. The point is r(0) = ⟨0, 0, 0⟩.
Q: A curve has r'(t) = ⟨3, 0, 4⟩ for all t. What is its curvature?
A: r''(t) = ⟨0, 0, 0⟩, so r' × r'' = ⟨0, 0, 0⟩. κ = 0. The curve is a straight line.
The tangent vector r'(t) becomes the velocity vector when t is time, leading directly into the motion problems covered later in Math 126 (acceleration, normal and tangential components). Curvature reappears in the study of surfaces, where Gaussian curvature and mean curvature generalise the idea to 2D. The cross-product curvature formula relies on the same determinant skills practised in the vectors and parallelogram notes.
Tags: vector-valued function, parametric curve, tangent line, tangent vector, derivative of vector function, curvature, kappa, maximum curvature, cross product curvature formula, space curve, Math 126, Calculus III, UW, multivariable calculus, arc length