Calculus Potluck – The Ultimate Buffet

Although the last Calculus Potluck was an epic fail, I do not want to give up this project. And this time, I am going to do something slightly different.

As it is indicated by the project name, I am going to keep posting practice problems until no one wants to solve them.

Here is how it works.

1. I will post 3 practice problems at a time.
2. Send me an email once you figure them out.
3. I will post solutions so that you can check your work.
4. Repeat if I receive emails from at least 3 students.

Practice Problems:

1. (a) Find the equation in the form $Ax+By+Cz = D$ of the plane $P$ which contains the line $L$ given by $x = 1-t, y = 1+2t, z = 2-3t$ and the point $(- 1, 1, 2)$. (b) Let $Q$ be the plane $2x+y+z = 4$. Find the component of a unit normal vector for $Q$ projected on a unit direction vector for the line $L$ of part(a).
2. Let $L$ denote the line which passes through $(0,0,1)$ and is parallel to the line in the $xy$-plane given by $y = 2x$. (a) Sketch $L$ and give its equation in vector-parametric form. (b) Let $P$ be the plane which passes through $(0,0,1)$ and is perpendicular to the line $L$ of part(a). Sketch in $P$ (above) and give its equation in point-normal form.
3. Let $r(t) = \langle \cos(e^t), \sin(e^t), e^t\rangle$. (a) Compute and simplify the unit tangent vector $T(t) = r'(t) / | r'(t) |$. (b) Compute $T'(t)$.
4. Consider the function $F(x,y,z)=z\sqrt{x^2+y}+2y/z$. (a) The point $P_0: (1,3,2)$ lies on the surface $F(x,y,z)=7$. Find the equation of the tangent plane to the surface $F(x,y,z)=7$ at $P_0$. (b) If starting at $P_0$ a small change were to be made in only one of the variables, which one would produce the largest change (in absolute value) in $F$? If the change of this variable was of size $0.1$, approximately how large would the change in $F$? (c) What distance from $P_0$ in the direction $\pm(-2,2,-1)$ will produce an approximate change in $F$ of size $0.1$ unites, according to the linearization of $F$?
5. Let $f(x,y)=x+4y+\frac{2}{xy}$. (a) Find the critical points of $f(x,y)$. (b) Use the second-derivative test to test the critical points found in part (a).
6. Let $P$ be the plane with equation $Ax+By+Cz=D$ and $P_0=(x_0, y_0, z_0)$ be a point which is not on $P$. Use the Lagrange multiplier method to set up the equations satisfied by the point $(x,y,z)$ on $P$ which is closest to $P_0$. (Do not solve.)
7. Let $F(x, y, z)$ be a smooth function of three variables for which $\nabla F(1, -1, \sqrt{2})=(1, 2, -2)$. Use the Chain Rule to evaluate $\partial F/\partial \phi$ at $(\rho,\phi,\theta) = (2, \pi/4,-\pi/4)$.
8. Suppose $\iint_R fdA = \int_0^2\int_{x^2}^{2\sqrt{2x}} f(x,y) dy dx$. (a) Sketch the region $R$. (b) Rewrite the double integral as an iterated integral with the order interchanged.
9. Let $G$ be the solid 3-D cone bounded by the lateral surface given by $z = 2 \sqrt{x^2 + y^2}$ and by the plane $z = 2$. The problem is to compute $\bar{z}$ which is the $z$-coordinate of the center of mass of $G$, in the case where the density is equal to the height above the $xy$-plane. (a) Find the mass of $G$ using cylindrical coordinates (b) Set up the calculation for $\bar{z}$ using cylindrical coordinates. (c) Set up the calculation for $\bar{z}$ using spherical coordinates.
10. $F(x,y,z)=(y+y^2z)i+(x-z+2xyz)j+(-y+xy^2)k$. (a) Show that $F(x,y,z)$ is a gradient field using the derivative conditions. (b) Find a potential function $f(x,y,z)$ for $F(x,y,z)$. (c) Find $\int_C F\cdot dr$, where $C$ is the straight line joining the points $(2,2,1)$ and $(1,-1,2)$.
11. In this problem $S$ is the surface given by the quarter of the right-circular cylinder centered on the $z$-axis, of radius 2 and height 4, which lies in the first octant. The field $F(x,y,z) = xi$. a) Sketch the surface $S$ and the field $F$. (b) Compute the flux integral (Use the normal which points ’outward’ from $S$, i.e. on the side away from the $z$-axis.) (c) $G$ be the 3D solid in the first octant given by the interior of the quarter cylinder defined above. Use the divergence theorem to compute the flux of the field $F = xi$ out of the region $G$.
12. $F(x, y, z) = (yz) i + (-xz) j + k$. Let $S$ be the portion of surface of the paraboloid $z=4-x^2-y^2$ which lies above the first octant; and let $C$ be the closed curve $C = C_1 + C_2 + C_3$, where the curves $C_1$, $C_2$ and $C3$ are the three curves formed by intersecting $S$ with the $xy$, $yz$ and $xz$ planes respectively (so that $C$ is the boundary of $S$). Orient $C$ so that it is traversed counter-clockwise when seen from above in the first octant. (a) Use Stokes’ Theorem to compute the loop integral $\oint_C F\cdot dr$ by using the surface integral over the capping surface $S$. (b) Set up and evaluate the loop integral $\oint_C F\cdot dr$ directly by parametrizing each piece of the curve $C$ and then adding up the three line integrals.

Solutions:

1. (a) The planes goes through $(1,1,2)$ and $(-1,1,2)$ and contains the direction of $L$, $(-1,2,-3)$. Therefore the normal vector of the plane is $(2,0,0)\times(-1,2,-3)=(0,6,4)$. Hence the equation of the plane is $6y+4z=6\times 1+4\times 2=14$ which boils down to $3y+2z=7$. (b) A unit normal vector for $Q$ is $u=(2,1,1)/\sqrt{6}=(2,1,1)/\sqrt{6}$(or alternatively $u=(-2,-1,-1)/\sqrt{6}$). On the other hand, a unit direction vector for the line $L$ is $v=(-1,2,-3)/\sqrt{14}$ (or alternatively $v=(1,-2,3)/\sqrt{14}$. Therefore the component $u$ on $v$ is $u\cdot v=\frac{-3}{2\sqrt{21}}$.
2. (a) Direction vector for $L$ is $v=(1,2,0)$ because the line in the $xy$-plane given by $y=2x$ can be parametrized as $x=t, y=2t, z=0$. The vector-parametric form of $L$ is $r=(t,2t,1)$. (b) Normal vector for $P$ is $(1,2,0)$ since the plane $P$ is perpendicular to the line $L$. The point-normal form of $P$ is $1(x-0)+2(y-0)+0(z-1)=0$ or $x+2y=0$.
3. (a) $r'(t) = (-\sin(e^t)e^t, \cos(e^t)e^t, e^t)$ implies $|r'(t)|=e^t\sqrt{2}$. Therefore $T(t)=\frac{1}{\sqrt{2}}(-\sin(e^t),\cos(e^t), 1)$. (b) $T'(t)=\frac{-e^t}{\sqrt{2}}(\cos(e^t), \sin(e^t), 0)$.
4. (a) $F_x=\frac{xz}{\sqrt{x^2+y}}$ implies $F_x(1,3,2)=1$. $F_y=\frac{z}{2\sqrt{x^2+y}}+\frac{2}{z}$ implies $F_y(1,3,2)=3/2$. $F_z=\sqrt{x^2+y}-\frac{2y}{z^2}$ implies $F_z(1,3,2)=1/2$. Therefore the normal vector is $(1,3/2,1/2)$ and the equation of the tangent plane is $1(x-1)+3/2(y-2)+1/2(z-2)=0$ or $2x+3y+z=13$. (b) At $P_0$, we have $|F_y|>|F_x|,|F_z|$. So a change in $y$ produces the largest change in $F$ and $\Delta F=F_y\Delta y=\frac{3}{2}\times(3.1-3)=0.15$. (c) The directional derivative of $F$ in the direction $s=(-2,2,-1)$ is $\frac{dF}{ds}=\frac{s}{|s|}\cdot \nabla F=\frac{(-2,2,-1)}{3}\cdot (1,3/2,1/2)=1/6$. Since we want $\Delta F = 0.1$ and we know $\Delta F = \frac{dF}{ds}\Delta s=\Delta s/6$, $\Delta s = 0.6$.
5. (a) Solve $f_x=1-2/(x^2)y=0, f_y=4-2/(xy^2)$ and get $x=2, y=1/2$. There is one critical point at $(2,1/2)$. (b) $f_xx=4/(x^3y), f_xy=2/(x^2y^2), f_yy=4/(xy^3)$, $A=f_{xx}(2,1/2)=1, B=f_{xy}(2,1/2)=2, C=f_{yy}(2,1/2)=16$. Therefore $AC-B^2=12$, $f$ has a local minimum at $(2,1/2)$.
6. The object is to minimize $f(x,y,z)=(x-x_0)^2+(y-y_0)^2+(z-z_0)^2$ subject to $g(x,y,z)=Ax+By+Cz=D$. The equations given by Lagrange multiplier method are $2(x-x_0)=\lambda A, 2(y-y_0)=\lambda B, 2(z-z_0)=\lambda C, Ax+By+Cz=D$.
7. Recall $x=\rho\sin\phi\cos\theta, y=\rho\sin\phi\sin\theta, z=\rho\cos\phi$ and $x_\phi=\rho\cos\phi\cos\theta, y_\phi=\rho\cos\phi\sin\theta, z_\phi=-\rho\sin\phi$. Therefore $(\rho, \phi, \theta)=(2,\pi/4,-\pi/4)$ represents the point $(1,-1,\sqrt{2})$. By the chain rule, $\begin{aligned} & \frac{\partial F}{\partial\phi}(2,\pi/4,-\pi/4) \\ = & \frac{\partial F}{\partial x}(1,-1,\sqrt{2})\frac{\partial x}{\partial \phi}(2,\pi/4,-\pi/4) \\ & + \frac{\partial F}{\partial y}(1,-1,\sqrt{2})\frac{\partial y}{\partial \phi}(2,\pi/4,-\pi/4) \\ + & \frac{\partial F}{\partial z}(1,-1,\sqrt{2})\frac{\partial z}{\partial \phi}(2,\pi/4,-\pi/4) \\ = & 1\times 1+2\times (-1)+(-2)\times (-\sqrt{2})=2\sqrt{2}-1.\end{aligned}$
8. (a) The region $R$ is enclosed by $y=x^2$ and $y=2\sqrt{2}x$. (b) Since $R$ is also the region described by $y^2/8\leq x\leq \sqrt{y}$ with $0\leq y\leq 4$. The double integral can be rewritten as $\int_0^4\int_{y^2/8}^{\sqrt{y}}f(x,y)dxdy$.
9. (a) Since the density of $G$ at point $(x,y,z)$ is $z$, the mass of $G$ is $\iiint_G z dV = \int_0^{2\pi}\int_0^1\int_{2r}^2 zdzrdrd\theta=\int_0^{2\pi}\int_0^12(1-r^2)rdrd\theta=\pi$. (b) $\bar{z}=\frac{1}{M}\iiint_G z^2 dV=\frac{1}{\pi}\int_0^{2\pi}\int_0^1\int_{2r}^2 z^2dzrdrd\theta$. (c) In spherical coordinates: $z=2$ becomes to $\rho\cos\theta=2 \implies \phi=2\sec\phi$. Limites on $G$ under the spherical coordinates: $0\leq\rho\leq 2\sec\phi, 0\leq\phi\leq\arctan(1/2), 0\leq\theta\leq 2\pi$. Therefore $\bar{z}=\frac{1}{M}\iiint_G z^2 dV=\frac{1}{\pi}\int_0^{2\pi}\int_0^{\arctan(1/2)}\int_0^{2\sec\phi}(\rho\cos\phi)^2\rho^2\sin\phi d\rho d\phi d\theta.$.
10. (a) We have $F=(P,Q,R)$, where $P=y+y^2z, Q=x-z+2xyz, R=-y+xy^2$. Check $\partial P/\partial z=y^2=\partial R/\partial x; \partial Q/\partial z=-1+2xy=\partial R/\partial y; \partial P/\partial y=1+2yz=\partial Q/\partial x$. (b) Suppose $f(x,y,z)$ is the potential function. Then $f_x = P = y+y^2z$ gives $f = xy+xy^2z+a(x,y)$. Then $f_y=x+2xyz+a_y=x-z+2xyz$. Therefore $a_y=-z$ implies $a=-yz+b(z)$ which implies $f=xy+xy^2z-yz+b(z)$. Then $f_z=xy^2-y+b'(z)=-y+xy^2$ implies $b(z)$ is a constant. So $f(x,y,z)=xy-yz+xy^2z+C$. (c) The work is the difference of the potential function at the end points, i.e., $f(1,-1,2)-f(2,2,1)=-10+3=-7$.
11. (b) We know $\hat{n}=\frac{1}{2}(x,y,0)$ and $F=(x,0,0)$. Thus $F\cdot \hat{n}=x^2/2$ and in cylindrical coordinates $dS=2dzd\theta$. On the surface, $x=2\cos\theta$ and the limites of the integration are $0\leq z\leq 4, 0\leq\theta\leq\pi/2$. So $\iint_S F\cdot\hat{n}dS=\iint_S x^2/2 dS=\int_0^{\pi/2}\int_0^4\frac{1}{2}(2\cos\theta)^2 2dzd\theta=4\pi$. (c) $div(F)=\nabla\cdot F=1$ implies $\iint_G\nabla\cdot F dV=\iiint_G 1dV=vol(G)=4\pi$.
12. (a) $\nabla\times F=(x,y,-2z)$ and $\hat{n}dS=(-z_x,-z_y,1)dA=(2x,2y,1)dA$. Therefore $(\nabla\times F)\cdot\hat{n}dS=(2x^2+2y^2-2z)dA=4(x^2+y^2-2)dA$. The limites of integration on $R$ are $0\leq r\leq 2, 0\leq\theta\leq\pi/2$. So $\iint_S(\nabla\times F)\cdot\hat{n}dS=\iint_R4(x^2+y^2-2)dA=4\int_0^{\pi/4}\int_0^24(r^2-2)rdrd\theta=0$. (b) $F=(yz,-xz,1)\implies \int_CF\cdot dr=\int_C(yz)dx-(xz)dy+1dz$. $C_1: x=0, y=t, z=4-t^2$ where t goes from 2 to 0. $\int_{C_1}F\cdot dr=\int_2^0(-2t)dt=4$. $C_2: x=t, y=0, z=4-t^2$ where $t$ goes from 0 to 2. $\int_{C_2}F\cdot dr=\int_0^2(-2t)dt=-4$. On $C_3$, $z=0, dz=0$. So $\int_{C_3}F\cdot dr=0$. Thus $\oint_C F\cdot dr=0$.