# 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.