Difficulty: Intermediate to Advanced | Prerequisites: Partial derivatives, linear approximation, vector operations
This set of notes covers four major topics from the MATH 215 midterm: parametric space curves (tangent lines, cusps, arc length), related rates via the multivariable chain rule, tangent planes and directional derivatives on implicit surfaces, and finding and classifying critical points using the second derivative test. Together they represent the computational core of the first half of Calculus III.
Parametric curve / space curve
A curve defined by r(t) = (x(t), y(t), z(t)). The parameter t traces out the path over an interval.
Tangent vector
r'(t) = (x'(t), y'(t), z'(t)). Points in the direction of motion along the curve at time t.
Cusp
A point where r'(t) = (0, 0, 0), so the tangent direction is undefined. The curve typically reverses or changes direction sharply.
Arc length
The total distance travelled along a curve: s = ∫ ||r'(t)|| dt. In simple terms, you integrate the speed.
Frustum (truncated cone)
A cone with the top sliced off, having two circular faces of different radii and a height. Its volume is V = (πh/3)(r₁² + r₁ r₂ + r₂²).
Level set / level surface
The set of points where a function equals a constant: g(x, y, z) = c. A surface defined implicitly.
Gradient
∇g = (g_x, g_y, g_z). Points in the direction of steepest increase and is perpendicular to the level surface at each point.
Directional derivative
D_u f = ∇f · û. The rate of change of f in the direction of the unit vector û.
Critical point
A point where ∇f = (0, 0). Candidates for local maxima, local minima, or saddle points.
Second derivative test (discriminant)
D = f_{xx} f_{yy} – (f_{xy})². If D > 0 and f_{xx} < 0 it is a local max; D > 0 and f_{xx} > 0 a local min; D < 0 a saddle.
Saddle point
A critical point where the function increases in some directions and decreases in others. Think of the centre of a horse's saddle.
Given r(t) = (sin³ t, cos³ t, sin² t) for 0 ≤ t ≤ π, find the tangent line at (1/(2√2), 1/(2√2), 1/2).
Identify the parameter value: t = π/4
Compute r'(t) = (3 sin² t cos t, −3 cos² t sin t, 2 sin t cos t)
Evaluate: r'(π/4) = (3/(2√2), −3/(2√2), 1)
Write the tangent line: l(s) = r(π/4) + s r'(π/4)
At the point (1, 0, 1), we need t = π/2. But r'(π/2) = (0, 0, 0). This means the tangent vector vanishes, which signals a cusp. Approaching π/2 from the left and right gives unit tangent vectors pointing in different directions, so no single tangent line exists at that point.
For the same curve on [0, π]:
||r'(t)|| = √13 |sin t cos t|
Split the integral at π/2 because sin t cos t changes sign:
s = √13 ∫₀^{π/2} sin t cos t dt – √13 ∫_{π/2}^π sin t cos t dt
Substituting u = sin t: both integrals evaluate to √13 × 1/2
Total arc length: s = √13
The absolute value is critical. Without it, the second half of the integral would cancel the first, giving zero.
A truncated cone (frustum) has larger radius r₁, smaller radius r₂, and height h. Its volume is:
V = (πh/3)(r₁² + r₁ r₂ + r₂²)
This formula comes from subtracting the small cone's volume from the large cone's volume and simplifying using similar triangles.
All three quantities change with time, so:
dV/dt = (∂V/∂r₁)(dr₁/dt) + (∂V/∂r₂)(dr₂/dt) + (∂V/∂h)(dh/dt)
The partial derivatives are:
∂V/∂r₁ = (2r₁ + r₂)πh / 3
∂V/∂r₂ = (r₁ + 2r₂)πh / 3
∂V/∂h = (r₁² + r₁ r₂ + r₂²)π / 3
Given r₁ = 15, r₂ = 10, h = 8, dr₁/dt = 2, dr₂/dt = −1, dh/dt = −3:
Plug in each term and sum. The result is dV/dt = –355π cm³/s.
The volume is decreasing, which makes sense: the height is shrinking fast and the smaller radius is also shrinking, even though the larger radius is growing.
Write down the geometric formula relating all variables
Differentiate with respect to time using the chain rule
Substitute the given values only after differentiating
Include units in the final answer
Given the surface S defined by g(x, y, z) = z²y² + xz – x² = −1, find the tangent plane at (2, 1, 1).
Compute the gradient: ∇g = (z – 2x, 2yz², 2zy² + x)
Evaluate at the point: ∇g(2, 1, 1) = (−3, 2, 4)
The tangent plane is: −3(x – 2) + 2(y – 1) + 4(z – 1) = 0
The gradient of the defining function g is always normal to the level surface g = c.
With f(x, y, z) = xy + z² at the same point:
∇f(2, 1, 1) = (1, 2, 2)
The unit normal from the tangent plane: n̂ = (−3, 2, 4)/√29
D_n f = (1, 2, 2) · (−3, 2, 4)/√29 = (−3 + 4 + 8)/√29 = 9/√29
If a curve C lies on S and f is constant along C, then C's tangent vector must be perpendicular to both ∇g (normal to S) and ∇f (normal to f's level surface). The cross product gives the direction:
v = ∇f × ∇g = (1, 2, 2) × (−3, 2, 4) = (4, −10, 8)
The tangent line is l(t) = (2, 1, 1) + t(4, −10, 8).
This is a powerful technique: whenever a curve lies on the intersection of two surfaces (or one surface with a constraint), its tangent direction is the cross product of the two gradients.
For f(x, y) = x²y – 2xy² – 3x² + 6xy:
∇f = (2(x – y)(y – 3), x(x – 4y + 6))
Set f_x = 0: either y = 3 or x = y
Set f_y = 0: either x = 0 or x – 4y + 6 = 0
Combine systematically, solving each pair:
(0, 0), (0, 3), (6, 3), (2, 2)
Compute the second partial derivatives:
f_{xx} = 2y – 6
f_{yy} = −4x
f_{xy} = 2x – 4y + 6
For each critical point, evaluate D = f_{xx} f_{yy} – (f_{xy})²:
(0, 0): f_{xx} = −6, f_{yy} = 0, f_{xy} = 6. D = 0 – 36 = −36 < 0. Saddle.
(0, 3): f_{xx} = 0, f_{yy} = 0, f_{xy} = −6. D = 0 – 36 = −36 < 0. Saddle.
(6, 3): f_{xx} = 0, f_{yy} = −24, f_{xy} = 6. D = 0 – 36 = −36 < 0. Saddle.
(2, 2): f_{xx} = −2, f_{yy} = −8, f_{xy} = 2. D = 16 – 4 = 12 > 0, and f_{xx} < 0. Local maximum.
On the entire plane, f has no absolute max or min. Along y = x, f(x, x) = –x³ + 3x², which diverges to ±∞.
On the triangle with vertices (0, 0), (0, 3), (6, 3): parametrise each boundary edge:
x = 0: f(0, y) = 0
y = 3: f(x, 3) = 3x² – 18x + 18x – 9x² + ... checking the exam, f ≡ 0 on all three sides
x = 2y: f(2y, y) = 0
So f = 0 on the entire boundary. The interior critical point (2, 2) lies inside the triangle with f(2, 2) = 4. The absolute max is 4 and the absolute min is 0.
\vec{\ell}(s) = \vec{r}(t_0) + s\,\vec{r}'(t_0) \quad \text{(tangent line)}s = \int_a^b \|\vec{r}'(t)\| \, dt \quad \text{(arc length)}\frac{dV}{dt} = \frac{\partial V}{\partial r_1}\frac{dr_1}{dt} + \frac{\partial V}{\partial r_2}\frac{dr_2}{dt} + \frac{\partial V}{\partial h}\frac{dh}{dt} \quad \text{(chain rule)}\nabla g \cdot (\vec{r} - \vec{r}_0) = 0 \quad \text{(tangent plane to level surface } g = c \text{)}D_{\hat{u}} f = \nabla f \cdot \hat{u} \quad \text{(directional derivative)}D = f_{xx}f_{yy} - (f_{xy})^2 \quad \text{(second derivative test discriminant)}Students forget to take the absolute value of ||r'(t)|| inside the arc length integral when the speed expression changes sign. Splitting the integral at the sign change is essential.
In related rates, a common mistake is plugging in numerical values before differentiating. Always differentiate first, then substitute.
When computing the tangent plane to an implicit surface, students sometimes use the gradient of the wrong function. If the surface is defined by g(x, y, z) = c, you need ∇g, not the gradient of some other function mentioned in the problem.
On the second derivative test, D = 0 is inconclusive, not a saddle. Students often guess "saddle" when D = 0; the test simply gives no information in that case.
⚠️ Tangent line problems for parametric curves can include cusp traps where r'(t₀) = 0. The exam tests whether you recognise this and explain the geometry.
⚠️ Related rates problems on MATH 215 exams typically involve three or more variables. The chain rule setup must account for every variable that changes with time.
⚠️ The "curve on a surface where f is constant" setup (tangent = cross product of two gradients) is a classic exam question that combines three concepts in one.
⚠️ For absolute extrema on a closed region, you must check the boundary completely. Parametrise every edge, not just the ones that look interesting.
What does it mean geometrically when r'(t₀) = (0, 0, 0)?
True or false: arc length can be negative.
Fill in the blank: the discriminant D = f_{xx} f_{yy} – (____)².
True or false: ∇g is tangent to the level surface g = c.
If D > 0 and f_{xx} > 0 at a critical point, that point is a ______.
Answers: 1. There is a cusp or the parametrisation stalls; the tangent line is undefined. 2. False (it is an integral of a non-negative quantity). 3. f_{xy}. 4. False (∇g is perpendicular to it). 5. Local minimum.
Q: Find the tangent line to r(t) = (t², t³, t) at t = 1.
A: r(1) = (1, 1, 1), r'(t) = (2t, 3t², 1), r'(1) = (2, 3, 1). Tangent line: l(s) = (1 + 2s, 1 + 3s, 1 + s).
Q: A sphere of radius r and a cylinder of radius R and height h are both changing. r increases at 2 cm/s, R decreases at 1 cm/s, h increases at 3 cm/s. When r = 5, R = 4, h = 10, at what rate is the total volume (sphere + cylinder) changing?
A: V = (4/3)πr³ + πR²h. dV/dt = 4πr²(dr/dt) + 2πRh(dR/dt) + πR²(dh/dt) = 4π(25)(2) + 2π(4)(10)(−1) + π(16)(3) = 200π – 80π + 48π = 168π cm³/s.
Q: Find the tangent plane to x² + y² + z² = 14 at (1, 2, 3).
A: g = x² + y² + z², ∇g = (2x, 2y, 2z). At (1, 2, 3): ∇g = (2, 4, 6). Tangent plane: 2(x – 1) + 4(y – 2) + 6(z – 3) = 0, or x + 2y + 3z = 14.
Q: Find and classify the critical points of f(x, y) = x² + y² – 4x + 6y.
A: f_x = 2x – 4 = 0, f_y = 2y + 6 = 0. Critical point: (2, −3). f_{xx} = 2, f_{yy} = 2, f_{xy} = 0. D = 4 > 0 and f_{xx} > 0. Local minimum.
Arc length calculations connect to curvature (κ = ||T'(s)||) studied later in the course. The gradient and directional derivative ideas extend to Lagrange multipliers for constrained optimisation. The second derivative test is the two-variable version of the concavity test from Calculus I, and it generalises to the bordered Hessian in constrained problems.
Related Terms / Search Tags
parametric curve, space curve, tangent line, tangent vector, cusp, arc length, frustum, truncated cone, related rates, chain rule, multivariable chain rule, level surface, implicit surface, gradient, normal vector, tangent plane, directional derivative, critical point, second derivative test, discriminant, saddle point, local maximum, local minimum, absolute extrema, closed region, boundary, MATH 215, Calculus III, University of Michigan, midterm 1