Source: Math 126 Spring 2023 Final Exam, University of Washington
Tags: vector functions, curve of intersection, curvature formula, cross product curvature, parametric curves 3D, MATH 126, Calculus III
Difficulty: Intermediate | Prerequisites: Parametric equations, derivatives, cross product
Vector-valued functions let you describe curves in 3D as a single object r(t) = ⟨x(t), y(t), z(t)⟩. Curvature measures how sharply a curve bends at each point. Both ideas come up constantly in physics (particle motion, electromagnetic fields) and in later calculus topics (line integrals, flux). You should be comfortable with differentiation and the cross product before tackling this material.
To find the curve where two surfaces meet, substitute one equation into the other and parametrise. Curvature at a point uses the formula κ = |r′ × r″| / |r′|³, which you evaluate by finding the right value of t first.
Vector function (vector-valued function)
A function r(t) = ⟨x(t), y(t), z(t)⟩ that outputs a vector for each value of t. Think of it as a set of instructions that traces out a curve in space as t changes.
Curve of intersection
The curve where two surfaces meet. You find it by combining the two surface equations to eliminate a variable, then parametrising what remains.
Curvature (κ)
A scalar that measures how fast a curve is changing direction at a given point. A straight line has curvature 0. A tight bend has large curvature. In simple terms, it tells you how "curvy" the curve is.
Unit tangent vector (T)
The normalised derivative r′(t)/|r′(t)|. It points in the direction of travel along the curve with magnitude 1.
Given two surface equations, substitute one into the other to reduce variables.
Look for forms you can parametrise with cos t and sin t (ellipses, circles).
Example: find the curve where 5x² + y² − z² = 4 meets z = x.
Substitute z = x: 5x² + y² − x² = 4, giving 4x² + y² = 4.
This is an ellipse. Set x = cos t, y = 2 sin t (so that 4cos²t + 4sin²t = 4).
Since z = x, we get z = cos t.
Vector function: r(t) = ⟨cos t, 2 sin t, cos t⟩.
The curvature formula is κ = |r′(t) × r″(t)| / |r′(t)|³.
Steps:
Find r′(t) and r″(t).
Determine the value of t that corresponds to the given point.
Evaluate r′ and r″ at that t.
Compute the cross product r′ × r″.
Take magnitudes and divide.
Example: r(t) = ⟨t³, t² − 1, 3t + 7⟩ at the point (−8, 3, 1).
r′(t) = ⟨3t², 2t, 3⟩, r″(t) = ⟨6t, 2, 0⟩.
At (−8, 3, 1): t³ = −8, so t = −2.
r′(−2) = ⟨12, −4, 3⟩, |r′(−2)| = √(144 + 16 + 9) = 13.
r″(−2) = ⟨−12, 2, 0⟩.
r′(−2) × r″(−2) = ⟨−6, −36, −24⟩, magnitude = 6√53.
κ = 6√53 / 13³ = 6√53 / 2197.
\kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}Students often forget to find the correct value of t before evaluating derivatives. The point (−8, 3, 1) does not mean t = −8. You must solve for t from one of the component equations.
The curvature formula requires the cross product, not the dot product. These are different operations with different outputs.
When parametrising curves of intersection, students sometimes forget to check that both surface equations are satisfied. Always verify your parametrisation works in both original equations.
Curvature is always non-negative (it is a magnitude divided by a magnitude). If you get a negative number, you have made an error.
⚠️ Curve-of-intersection questions test whether you can combine surface equations and spot standard parametrisations (circles, ellipses).
⚠️ Curvature computation is heavily tested. The cross product and magnitude calculations are where most arithmetic errors occur, so check your work.
⚠️ Finding the correct t value from a given point is a common exam trap. Solve carefully.
True or False: The curvature of a straight line is 1. (False, it is 0.)
Fill in the blank: To parametrise the ellipse 4x² + y² = 4, set x = cos t and y = ______. (2 sin t)
True or False: If r(t) = ⟨t³, t² − 1, 3t + 7⟩ and the point is (−8, 3, 1), then t = −8. (False, t = −2.)
Q: Find a vector function for the curve of intersection of 5x² + y² − z² = 4 and z = x.
A: Substitute z = x to get 4x² + y² = 4. Set x = cos t, y = 2 sin t, z = cos t. So r(t) = ⟨cos t, 2 sin t, cos t⟩.
Q: Find the curvature of r(t) = ⟨t³, t² − 1, 3t + 7⟩ at (−8, 3, 1).
A: t = −2. r′(−2) = ⟨12, −4, 3⟩, r″(−2) = ⟨−12, 2, 0⟩. Cross product magnitude = 6√53. |r′| = 13. κ = 6√53/2197.
Vector functions connect directly to line integrals in Chapter 16, where you integrate a function along a curve parametrised by r(t). Curvature also appears in physics when analysing the centripetal acceleration of a moving particle.
vector-valued function, space curve, curvature formula, curve of intersection, parametric curve 3D, cross product curvature, unit tangent vector, r prime cross r double prime, MATH 126, Calculus III, UW Math 126