Problem Set | Mathematics - Wyatt's Notes

$f_x = \frac{2x}{x^2+y^2} + ze^{xz}$, $f_y = \frac{2y}{x^2+y^2}$, $f_z = xe^{xz}$.

At $(1,0,0)$: $f_x = 2 + 0 = 2$, $f_y = 0$, $f_z = 1$.

$\nabla f(1,0,0) = (2, 0, 1)$.

If you get this wrong, revise: Section 1.4 The Gradient.

$f_x = 3x^2 - 3y^2 = 0$ and $f_y = -6xy + 3y^2 = 3y(-2x + y) = 0$.

From $f_x = 0$: $x^2 = y^2$ So $y = \pm x$.

If $y = x$: $f_y = 3x(-2x + x) = -3x^2 = 0$ So $x = 0$. Point: $(0,0)$.

If $y = -x$: $f_y = 3(-x)(2x + x) = -9x^2 = 0$ So $x = 0$. Point: $(0,0)$.

The only critical point is $(0, 0)$. Now $f_{xx} = 6x$, $f_{yy} = -6x + 6y$, $f_{xy} = -6y$.

At $(0,0)$: $D = 0 \cdot 0 - 0 = 0$. The second derivative test is inconclusive.

To classify, note $f(x, y) = x^3 - 3xy^2 + y^3$. Along $y = 0$: $f(x, 0) = x^3$Which changes sign At $0$. Along $x = y$: $f(x, x) = -x^3$Which also changes sign but with opposite sign. Since the behaviour differs by direction, $(0, 0)$ is a saddle point.

If you get this wrong, revise: Section 4.2 Second Derivative Test.

Normalise: $\lVert \mathbf{v} \rVert = \sqrt{2}$ So $\mathbf{u} = (1/\sqrt{2},\, 1/\sqrt{2})$.

$f_x = e^x \cos y$, $f_y = -e^x \sin y$.

$\nabla f(0, \pi/2) = (e^0 \cos(\pi/2),\, -e^0 \sin(\pi/2)) = (0, -1)$.

$D_{\mathbf{u}} f = (0, -1) \cdot (1/\sqrt{2},\, 1/\sqrt{2}) = -\frac{1}{\sqrt{2}}$

If you get this wrong, revise: Section 1.5 Directional Derivatives.

Let $F(x,y,z) = x^2 z + y^2 z^2 - 5$. Then $F_x = 2xz$, $F_y = 2yz^2$, $F_z = x^2 + 2y^2 z$.

At $(1,1,1)$: $F_x = 2$, $F_z = 1 + 2 = 3$.

$\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{2}{3}$

If you get this wrong, revise: Section 1.9 Implicit Differentiation.

$f(0,0) = 0$, $f_x = \cos(x+y)$, $f_y = \cos(x+y)$ So $f_x(0,0) = f_y(0,0) = 1$.

$f_{xx} = -\sin(x+y)$, $f_{xy} = -\sin(x+y)$, $f_{yy} = -\sin(x+y)$ So $f_{xx}(0,0) = f_{xy}(0,0) = f_{yy}(0,0) = 0$.

$f(x,y) = 0 + x + y + \frac{1}{2}(0 \cdot x^2 + 2 \cdot 0 \cdot xy + 0 \cdot y^2) + R_2 = x + y + R_2$

Where $R_2 = O(\lvert x \rvert^3 + \lvert y \rvert^3)$.

If you get this wrong, revise: Section 1.10 Taylor”s Theorem.

The curves intersect when $x = x^2$I.e., $x(x-1) = 0$ So $x = 0$ and $x = 1$. For $x \in (0,1)$ $x^2 \lt x$ So $D = \\{(x,y) : 0 \leq x \leq 1,\, x^2 \leq y \leq x\\}$.

$\iint_D (x + y)\, dA = \int_0^1 \int_{x^2}^x (x + y)\, dy\, dx = \int_0^1 \left[xy + \frac{y^2}{2}\right]_{x^2}^x\, dx$

$= \int_0^1 \left(x^2 + \frac{x^2}{2} - x^3 - \frac{x^4}{2}\right)\, dx = \int_0^1 \left(\frac{3x^2}{2} - x^3 - \frac{x^4}{2}\right)\, dx$

$= \left[\frac{x^3}{2} - \frac{x^4}{4} - \frac{x^5}{10}\right]_0^1 = \frac{1}{2} - \frac{1}{4} - \frac{1}{10} = \frac{10 - 5 - 2}{20} = \frac{3}{20}$

If you get this wrong, revise: Section 2.2 General Regions.

$E = \\{(x,y,z) : 0 \leq x \leq 1,\, 0 \leq y \leq 1-x,\, 0 \leq z \leq 1-x-y\\}$.

$\iiint_E x\, dV = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} x\, dz\, dy\, dx = \int_0^1 \int_0^{1-x} x(1-x-y)\, dy\, dx$

$= \int_0^1 x\left[(1-x)y - \frac{y^2}{2}\right]_0^{1-x}\, dx = \int_0^1 x \cdot \frac{(1-x)^2}{2}\, dx$

$= \frac{1}{2}\int_0^1 x(1 - 2x + x^2)\, dx = \frac{1}{2}\int_0^1 (x - 2x^2 + x^3)\, dx$

$= \frac{1}{2}\left[\frac{x^2}{2} - \frac{2x^3}{3} + \frac{x^4}{4}\right]_0^1 = \frac{1}{2}\left[\frac{1}{2} - \frac{2}{3} + \frac{1}{4}\right] = \frac{1}{2} \cdot \frac{6 - 8 + 3}{12} = \frac{1}{24}$

If you get this wrong, revise: Section 2.3 Triple Integrals.

Use polar coordinates: $1 \leq r \leq 2$, $0 \leq \theta \leq 2\pi$.

$\iint_D e^{x^2+y^2}\, dA = \int_0^{2\pi} \int_1^2 e^{r^2}\, r\, dr\, d\theta = 2\pi \int_1^2 r e^{r^2}\, dr$

Let $u = r^2$, $du = 2r\, dr$:

$= 2\pi \cdot \frac{1}{2}\int_1^4 e^u\, du = \pi(e^4 - e)$

If you get this wrong, revise: Section 2.4 Change of Variables.

Use cylindrical coordinates. The cone $z = r$ intersects $z = 1$ at $r = 1$. $E' = \\{(r, \theta, z) : 0 \leq r \leq 1,\, 0 \leq \theta \leq 2\pi,\, r \leq z \leq 1\\}$.

$\iiint_E z\, dV = \int_0^{2\pi} \int_0^1 \int_r^1 z\, r\, dz\, dr\, d\theta = 2\pi \int_0^1 r\left[\frac{z^2}{2}\right]_r^1\, dr$

$= 2\pi \int_0^1 \frac{r}{2}(1 - r^2)\, dr = \pi \int_0^1 (r - r^3)\, dr = \pi\left[\frac{r^2}{2} - \frac{r^4}{4}\right]_0^1 = \pi \cdot \frac{1}{4} = \frac{\pi}{4}$

If you get this wrong, revise: Section 2.5 Coordinate System Worked Examples.

$P = 3y - e^{\sin x}$, $Q = 7x + \sqrt{y^4 + 1}$.

$\frac{\partial Q}{\partial x} = 7, \quad \frac{\partial P}{\partial y} = 3$

By Green’s theorem:

$\oint_C P\, dx + Q\, dy = \iint_D (7 - 3)\, dA = 4 \cdot \pi \cdot 9 = 36\pi$

If you get this wrong, revise: Section 3.3 Green’s Theorem.

Curl:

$\nabla \times \mathbf{F} = \left(\frac{\partial (xy)}{\partial y} - \frac{\partial (xz)}{\partial z},\, \frac{\partial (yz)}{\partial z} - \frac{\partial (xy)}{\partial x},\, \frac{\partial (xz)}{\partial x} - \frac{\partial (yz)}{\partial y}\right)$

$= (x - x,\, y - y,\, z - z) = \mathbf{0}$

Divergence:

$\nabla \cdot \mathbf{F} = \frac{\partial (yz)}{\partial x} + \frac{\partial (xz)}{\partial y} + \frac{\partial (xy)}{\partial z} = 0 + 0 + 0 = 0$

Since the curl is zero and the domain is connected, $\mathbf{F}$ is conservative. Indeed, $\mathbf{F} = \nabla(xyz)$.

If you get this wrong, revise: Section 3.4 Curl and Divergence.

Take $S$ to be the disk $x^2 + y^2 \leq 1$, $z = 1$ with upward normal $\mathbf{n} = (0, 0, 1)$.

$\nabla \times \mathbf{F} = \left(\frac{\partial x}{\partial y} - \frac{\partial (-z)}{\partial z},\, \frac{\partial (2y)}{\partial z} - \frac{\partial x}{\partial x},\, \frac{\partial (-z)}{\partial x} - \frac{\partial (2y)}{\partial y}\right)$

$= (0 - (-1),\, 0 - 1,\, 0 - 2) = (1, -1, -2)$

$(\nabla \times \mathbf{F}) \cdot \mathbf{n} = (1, -1, -2) \cdot (0, 0, 1) = -2$

$\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (-2)\, dS = -2 \cdot \pi \cdot 1^2 = -2\pi$

If you get this wrong, revise: Section 3.5 Stokes’ Theorem.

$\nabla \cdot \mathbf{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 3$

$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E 3\, dV = 3 \cdot 1^3 = 3$

If you get this wrong, revise: Section 3.6 Divergence Theorem.

First check: $\nabla \times \mathbf{F} = (2 - 2,\, 0 - 0,\, 1 - 1) = \mathbf{0}$. Conservative.

$\frac{\partial \phi}{\partial x} = 2x + y \implies \phi = x^2 + xy + g(y,z)$

$\frac{\partial \phi}{\partial y} = x + g_y = x + 2z \implies g_y = 2z \implies g = 2yz + h(z)$

$\frac{\partial \phi}{\partial z} = 2y + h'(z) = 2y \implies h'(z) = 0 \implies h(z) = C$

$\phi(x,y,z) = x^2 + xy + 2yz + C$

If you get this wrong, revise: Section 3.7 Conservative Fields and Potential Functions.

Parametrise the cylinder: $\mathbf{r}(\theta, z) = (2\cos\theta,\, 2\sin\theta,\, z)$ for $0 \leq \theta \leq 2\pi$, $0 \leq z \leq 3$.

$\mathbf{r}_\theta = (-2\sin\theta,\, 2\cos\theta,\, 0)$, $\mathbf{r}_z = (0,\, 0,\, 1)$.

$\mathbf{r}_\theta \times \mathbf{r}_z = (2\cos\theta,\, 2\sin\theta,\, 0)$

$\lVert \mathbf{r}_\theta \times \mathbf{r}_z \rVert = \sqrt{4\cos^2\theta + 4\sin^2\theta} = 2$

On $S$: $x^2 + y^2 = 4$.

$\iint_S (x^2 + y^2)\, dS = \int_0^{2\pi} \int_0^3 4 \cdot 2\, dz\, d\theta = 8 \cdot 3 \cdot 2\pi = 48\pi$

If you get this wrong, revise: Section 5.5 Surface Integrals.

By Green’s theorem with $P = -y/2$ and $Q = x/2$:

$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1$

So the area is:

$A = \iint_D 1\, dA = \oint_C -\frac{y}{2}\, dx + \frac{x}{2}\, dy = \frac{1}{2}\oint_C x\, dy - y\, dx$

Parametrise the ellipse: $x = a\cos t$, $y = b\sin t$, $0 \leq t \leq 2\pi$.

$A = \frac{1}{2}\int_0^{2\pi} \left[a\cos t \cdot b\cos t - b\sin t \cdot (-a\sin t)\right]\, dt$

$= \frac{1}{2}\int_0^{2\pi} (ab\cos^2 t + ab\sin^2 t)\, dt = \frac{ab}{2}\int_0^{2\pi} 1\, dt = \pi ab$

If you get this wrong, revise: Section 3.3 Green’s Theorem.

$\nabla f = (2x, 2y, 2z)$, $\nabla g = (1, 1, -1)$ where $g = x + y - z - 1$.

$2x = \lambda$, $2y = \lambda$, $2z = -\lambda$ So $x = y = -z$.

From $x + y - z = 1$: $2x - (-x) = 3x = 1$ So $x = 1/3$, $y = 1/3$, $z = -1/3$.

$f(1/3, 1/3, -1/3) = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{1}{3}$

This is the minimum (the Hessian of $f$ is positive definite, and the constraint set is unbounded But $f \geq 0$).

If you get this wrong, revise: Section 4.3 Lagrange Multipliers.

$\mathbf{r}'(t) = (2t,\, 2,\, 1/t)$ So $\lVert \mathbf{r}'(t) \rVert = \sqrt{4t^2 + 4 + 1/t^2}$.

Note: $4t^2 + 4 + t^{-2} = (2t + 1/t)^2$. So $\lVert \mathbf{r}' \rVert = 2t + 1/t$.

$L = \int_1^e \left(2t + \frac{1}{t}\right)\, dt = \left[t^2 + \ln t\right]_1^e = e^2 + 1 - 1 - 0 = e^2$

If you get this wrong, revise: Section 5.1 Parametric Curves.

$\mathbf{r}'(t) = (1,\, 2t,\, 3t^2)$, $\mathbf{r}''(t) = (0,\, 2,\, 6t)$.

At $t = 1$: $\mathbf{r}' = (1, 2, 3)$, $\mathbf{r}'' = (0, 2, 6)$.

$\lVert \mathbf{r}' \rVert = \sqrt{1 + 4 + 9} = \sqrt{14}$.

$\mathbf{r}' \times \mathbf{r}'' = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 0 & 2 & 6 \end{vmatrix} = (12 - 6,\, -(6 - 0),\, 2 - 0) = (6, -6, 2)$

$\lVert \mathbf{r}' \times \mathbf{r}'' \rVert = \sqrt{36 + 36 + 4} = \sqrt{76} = 2\sqrt{19}$

$\kappa = \frac{2\sqrt{19}}{(\sqrt{14})^3} = \frac{2\sqrt{19}}{14\sqrt{14}} = \frac{\sqrt{266}}{98}$

If you get this wrong, revise: Section 5.2 Curvature and Torsion.

Use spherical coordinates. The sphere has $\rho = 2$. The plane $z = 1$ intersects when $2\cos\phi = 1$ So $\cos\phi = 1/2$Giving $\phi = \pi/3$.

The region: $0 \leq \rho \leq 2$, $0 \leq \phi \leq \pi/3$, $0 \leq \theta \leq 2\pi$.

$A = \int_0^{2\pi} \int_0^{\pi/3} \rho^2 \sin\phi\, d\phi\, d\theta = 4 \cdot 2\pi \int_0^{\pi/3} \sin\phi\, d\phi$

$= 8\pi \left[-\cos\phi\right]_0^{\pi/3} = 8\pi \left(-\frac{1}{2} + 1\right) = 8\pi \cdot \frac{1}{2} = 4\pi$

If you get this wrong, revise: Section 5.4 Surface Area.

Check: $\frac{\partial P}{\partial y} = e^{xy} + xye^{xy}$, $\frac{\partial Q}{\partial x} = e^{xy} + xye^{xy}$. These are equal, so $\mathbf{F}$ is conservative (on $\mathbb{R}^2$Which is connected).

Find $\phi$:

$\frac{\partial \phi}{\partial x} = ye^{xy} + 2x \implies \phi = e^{xy} + x^2 + g(y)$

$\frac{\partial \phi}{\partial y} = xe^{xy} + g'(y) = xe^{xy} + 2y \implies g'(y) = 2y \implies g(y) = y^2 + C$

$\phi(x,y) = e^{xy} + x^2 + y^2$

$\int_C \mathbf{F} \cdot d\mathbf{r} = \phi(1,1) - \phi(0,0) = (e + 1 + 1) - (1 + 0 + 0) = e + 1$

If you get this wrong, revise: Section 3.2 Line Integrals and Section 3.7 Conservative Fields.

$\mathbf{r}'(t) = (\sinh t,\, \cosh t,\, 1)$, $\mathbf{r}''(t) = (\cosh t,\, \sinh t,\, 0)$ $\mathbf{r}^{\prime\prime\prime}(t) = (\sinh t,\, \cosh t,\, 0)$.

At $t = 0$: $\mathbf{r}' = (0, 1, 1)$, $\mathbf{r}'' = (1, 0, 0)$ $\mathbf{r}^{\prime\prime\prime} = (0, 1, 0)$.

$\mathbf{r}' \times \mathbf{r}'' = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{vmatrix} = (0,\, 1,\, -1)$

$\lVert \mathbf{r}' \times \mathbf{r}'' \rVert = \sqrt{2}$

$(\mathbf{r}' \times \mathbf{r}'') \cdot \mathbf{r}^{\prime\prime\prime} = 1$

$\tau = \frac{1}{(\sqrt{2})^2} = \frac{1}{2}$

If you get this wrong, revise: Section 5.2 Curvature and Torsion.

Use spherical coordinates. The integrand is $\frac{1}{\rho}$.

$\iiint_E \frac{1}{\rho}\, dV = \int_0^{2\pi} \int_0^{\pi} \int_0^1 \frac{1}{\rho} \cdot \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta$

$= \left(\int_0^1 \rho\, d\rho\right)\left(\int_0^{\pi} \sin\phi\, d\phi\right)\left(\int_0^{2\pi} d\theta\right)$

$= \frac{1}{2} \cdot 2 \cdot 2\pi = 2\pi$

If you get this wrong, revise: Section 2.5 Coordinate System Worked Examples.