Practice Exam 3
Partial Derivatives
For \(f(x,y) = \ln(9 - x^2 - y^2)\), find the domain and sketch the level curves for \(f = 0\), \(f = \ln 5\) and \(f = 2\).
Solution
Domain: \(9 - x^2 - y^2 > 0\), the open disk \(x^2 + y^2 < 9\).
Level curves: \(\ln(9 - r^2) = k\Rightarrow r^2 = 9 - e^k\). Circles centred at the origin:
- \(k = 0\): \(r^2 = 8\), \(r\approx 2.83\).
- \(k = \ln5\): \(r^2 = 4\), \(r = 2\).
- \(k = 2\): \(r^2 = 9 - e^2\approx 1.61\), \(r\approx 1.27\).
Higher values sit on smaller circles — a peak at the origin (\(f(0,0) = \ln 9\approx 2.2\)) falling to \(-\infty\) at the boundary. The circles crowd together near \(r = 3\): the surface gets steep there.
Decide whether each limit exists as \((x,y)\to(0,0)\), and find it if so: (a) \(\dfrac{x^2 - y^2}{x^2 + y^2}\); (b) \(\dfrac{x^2 y}{x^2 + y^2}\).
Solution
(a) Along the \(x\)-axis (\(y = 0\)): \(x^2/x^2 = 1\). Along the \(y\)-axis (\(x = 0\)): \(-y^2/y^2 = -1\). Two paths, two values — the limit does not exist.
(b) Polar: \(x = r\cos\theta\), \(y = r\sin\theta\):
Bounded by \(r\) regardless of \(\theta\), so the limit exists and equals 0. The extra power of \(r\) in the numerator is what makes the difference between (a) and (b).
For \(f(x,y) = x^3y^2 - 3xy^4 + e^{2x}y\), find \(f_x\), \(f_y\), \(f_{xx}\), \(f_{xy}\) and \(f_{yx}\). Confirm Clairaut's theorem.
Solution
\(f_{xy} = f_{yx}\) ✓. When computing \(f_x\), the term \(-3xy^4\) is linear in \(x\) — its \(x\)-derivative is \(-3y^4\), with the \(y^4\) untouched. Treat \(y\) as a number and it is Calculus I.
Find the tangent plane to \(z = \sqrt{x^2 + y^2 + 1}\) at \((2, 2, 3)\), and use it to estimate \(f(2.1, 1.9)\). Compare with the true value.
Solution
True value: \(\sqrt{4.41 + 3.61 + 1} = \sqrt{9.02}\approx 3.0033\). The linearisation gave exactly 3 because the two displacements cancel along the tangent plane; the error \(0.0033\) is the second-order curvature the plane cannot see.
A resistor dissipates \(P = V^2/R\). The supply is \(V = 12.0\pm0.2\) V and the resistor is \(R = 6.0\pm0.3\ \Omega\). Estimate the worst-case uncertainty in \(P\), absolute and relative.
Solution
Worst case adds the absolute values — the signs of \(dV\) and \(dR\) are unknown. In relative form: \(\dfrac{dP}{P} = 2\dfrac{dV}{V} + \dfrac{dR}{R} = 2(1.67\%) + 5\% = 8.3\%\). A squared variable contributes twice its own relative tolerance — worth remembering for any power calculation.
(a) If \(z = x^2 + xy\), \(x = s + 2t\), \(y = st\), find \(\partial z/\partial t\) at \((s,t) = (1,1)\). (b) If \(x^3 + y^3 + z^3 + 6xyz = 1\) defines \(z\) implicitly, find \(\partial z/\partial x\).
Solution
(a) At \((1,1)\): \(x = 3\), \(y = 1\).
(b) With \(F = x^3 + y^3 + z^3 + 6xyz - 1\):
Valid wherever \(F_z\ne0\). The minus sign is the one people lose.
The potential in a region is \(V(x,y,z) = 5x^2 - 3xy + xyz\) volts (metres). At the point \((1, 2, 3)\): (a) find \(\nabla V\); (b) find \(\mathbf{E} = -\nabla V\) and \(|\mathbf{E}|\); (c) find the rate of change of \(V\) in the direction \(\langle 1, 1, -1\rangle\); (d) give the direction in which \(V\) decreases fastest.
Solution
(c) Unit vector \(\mathbf{u} = \langle 1, 1, -1\rangle/\sqrt3\):
(d) Fastest decrease is along \(-\nabla V\), i.e. along \(\mathbf{E}\) itself: \(\langle -10, 0, -2\rangle/\sqrt{104}\). A positive charge released here accelerates that way, and the potential drops at \(10.2\) V per metre along it.
Find and classify all critical points of \(f(x,y) = x^3 + y^3 - 3xy\).
Solution
Substituting, \(x = x^4\), so \(x = 0\) or \(x = 1\). Critical points: \((0,0)\) and \((1,1)\).
- \((0,0)\): \(D = -9 < 0\) → saddle point, \(f = 0\).
- \((1,1)\): \(D = 27 > 0\), \(f_{xx} = 6 > 0\) → local minimum, \(f = -1\).
\(f_{xy}\) is a constant here but it still enters \(D\) — \(D = f_{xx}f_{yy} - f_{xy}^2\), not \(f_{xx}f_{yy}\). Forgetting the \(-9\) would wrongly make the origin inconclusive.
Find the absolute maximum and minimum of \(f(x,y) = x^2 + 2y^2 - 2x\) on the closed disk \(x^2 + y^2\le4\).
Solution
Interior: \(f_x = 2x - 2 = 0\), \(f_y = 4y = 0\) → \((1, 0)\), inside the disk. \(f(1,0) = -1\).
Boundary: \(x = 2\cos t\), \(y = 2\sin t\):
Let \(c = \cos t\in[-1,1]\): \(h(c) = 8 - 4c^2 - 4c\), \(h'(c) = -8c - 4 = 0\) at \(c = -\tfrac12\), where \(h = 8 - 1 + 2 = 9\). Endpoints: \(h(1) = 0\), \(h(-1) = 8\).
Compare everything: \(-1,\ 9,\ 0,\ 8\). Absolute minimum \(-1\) at \((1,0)\); absolute maximum \(9\) at \(\cos t = -\tfrac12\), i.e. \((-1, \pm\sqrt3)\). Check: \(f(-1,\sqrt3) = 1 + 6 + 2 = 9\) ✓.
A total current \(I = 7\) A splits among three parallel resistors \(R_1 = 1\ \Omega\), \(R_2 = 2\ \Omega\), \(R_3 = 4\ \Omega\). Nature chooses the split that minimises the dissipated power \(P = I_1^2R_1 + I_2^2R_2 + I_3^2R_3\) subject to \(I_1 + I_2 + I_3 = 7\). Use Lagrange multipliers to find the currents, and interpret the result.
Solution
Interpretation: \(I_kR_k = \lambda/2 = 4\) V for every branch — the voltages across the parallel resistors are equal. That is Kirchhoff's voltage law, obtained here from a minimum-power principle rather than assumed. The power is \(16 + 8 + 4 = 28\) W, and \(\lambda = 8 = 2V\) is the marginal cost: each extra ampere of total current costs \(2V\) more watts at the margin.
Scoring
| Score | Where you are |
|---|---|
| 90–100 | Solid. On to multiple integrals. |
| 75–89 | Concepts fine; slips in \(f_{xy}^2\) or in normalising the direction vector. Check \(D\) includes the cross term and \(\mathbf{u}\) has length 1. |
| 60–74 | Boundary of the region skipped (Q9) or gradient direction confused with \(\mathbf{E}\) (Q7). Re-read Sections 6–7 with the labs open. |
| < 60 | Start from Section 3: partial differentiation must be automatic before the gradient can mean anything. |
Not normalising the direction vector (Q7c); dropping \(f_{xy}^2\) from \(D\) (Q8); forgetting the boundary or its endpoints (Q9); and the sign in \(\partial z/\partial x = -F_x/F_z\) (Q6b). The tolerance question (Q5) is lost by subtracting the two contributions instead of adding their magnitudes.