Extra Problems for Midterm 2


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1 Extra Problems for Midterm Sudesh Kalyanswamy Exercise (Surfaces). Find the equation of, and classify, the surface S consisting of all points equidistant from (0,, 0) and (,, ). Solution. Let P (x, y, z) be on the surface. The distance from P to (0,, 0) is x + (y + ) + z, and the distance from P to (,, ) is (x ) + (y ) + (z ). Therefore x + (y + ) + z = (x ) + (y ) + (z ). Square both sides and expand. x + y + z =, which is a plane. After simplifying, you should get 4x + 8y + 4x = 8, or Exercise (Surfaces). Classify the quadric surface given by the equation x + y x 4y z 5 = 0. Solution. Complete the square in the x and y to get something like which is an elliptic paraboloid. (x ) + (y ) = z = z + 8, Exercise 3 (Surfaces). Find the equation for the surface obtained by rotating y = x about the yaxis, and classify the surface. Solution 3. The equation is y = x + z, and this is an elliptic paraboloid. Exercise 4 (Surfaces). Find an equation of, and clasify, the surface consisting of all points equidistant from (, 0, 0) and x =. Solution 4. This is similar to the first exercise. The distance from a point P (x, y, z) on the surface to (, 0, 0) is (x + ) + y + z, and the distance to the plane x = is x. Therefore x = (x + ) + y + z. Square both sides and simplify to get 4x = y + z, which is, again, an elliptic paraboloid. Exercise 5 (Surfaces). Find the equation for, and classify, the surface obtained by rotating x = 3y about the xaxis. Solution 5. For a fixed xvalue, say x = a, the trace of the surface is a circle with equation y + z = ( a 3) = a 9. Therefore the surface will have equation y + z = x, which is a cone. 9 Exercise 6 (Level Curves and other stuff). Consider f(x, y) = ln(x y).
2 (a) Find and sketch the domain of f(x, y). (b) Find the range of f(x, y). (c) What do level curves of f(x, y) look like? (d) Find the tangent plane to f(x, y) at (, 0). (e) Approximate ln(. +.03). Solution 6. The domain is {(x, y) R : x y > 0}. The range is all of R. The level curves are parabolas: c = ln(x y) = e c = x y = y = x e c. For the tangent plane, we need our two partials: f x = x x y, f y = x y, so f x (, 0) = and f y (, 0) =. Since f(, 0) = ln( 0) = 0, the tangent plane is (x ) (y 0) (z 0) = 0 = z = (x ) y. Finally, to get the approximation, note that this value is what you get when you plug in x =. and y =.03 into the function, so the approximate value is what you get when you plug in these into the tangent plane: (. ) Exercise 7 (Level Curves/Tangent Plane). Let f(x, y) = x + y. (a) Plot the level curve to c = 0. (b) What is lim f(x, y)? (x,y) (0, ) (c) Find the tangent plane to f(x, y) at P (,, 3). Solution 7. (a) The graph of x + y = 0 is just the point (0, 0). (b) f(x, y) is continuous, so just plug in (0, ): f(0, ) =, so the limit is. (c) Near P, f(x, y) looks like x y by definition of absolute values. Therefore z = x y near P, which is already a plane. So the tangent plane will just be the function itself: x y z = 0. Exercise 8 (Partials). If f(x, y, z) = xy z + sin (x z), find f xzy. Solution 8. By Clairaut s theorem, since this function will have sufficiently nice partials, we can do this in any order. Notice that f y = xyz 3, so f yx = yz 3, and finally f yxz = 6yz. Exercise 9 (Partials). Does there exist f(x, y) such that f x = x + 4y and f y = 3x y. Solution 9. No, since f xy = 4, which does not equal f yx = 3.
3 Exercise 0 (Partials). The ellipsoid 4x + y + z = 6 intersects the plane y = in what shape? Find the tangent line to this shape at (,, ). Solution 0. It intersects it in an ellipse: 4x + + z = 6, so 4x + z = 8, which is an ellipse. There are several ways of going about getting the line. We need a direction vector, so one could parametrize this ellipse: x = cos(t), z = sin(t). The point (,, ) corresponds to t = π/4. Since x (π/4) = and z (π/4) =, a direction vector is, 0,. So the tangent line is,, + t, 0,. Exercise (Partials). If f(x, y) = x(x + y ) 3/ e sin(x y), find f x (, 0). Solution. Notice that f(x, 0) = x(x ) 3/ = x. Therefore f x (, 0) = x 3 x= =. Exercise (Partials). Does there exist f(x, y) such that f x = x and f y = y? Solution. Observe that f xy = f yx = 0, so maybe. If f x = y, then integrating with respect to x gives f(x, y) = x + g(y) for some function g(y). Taking the partial of this with respect to y gives f y = g (y) = y, so one candidate for g(y) is y3. Observe, then, that 3 works. f(x, y) = x + y3 3 Exercise 3 (Tangent Plane). Show that every tangent plane to the cone x +y = z passes through the origin. Solution 3. Let P (a, b, c) be a point on the cone. The tangent plane to P is a(x a) + b(x b) c(x c) = 0. If you plug in (0, 0, 0) and recall that a + b = c since P is on the cone, you observe that the origin is on the plane, as desired. Exercise 4 (Tangent Plane). A surface contains the curves r (t) = t, t, t 3 and r (s) =, sin(s), cos(s). Find the tangent plane to (,, ). Solution 4. Notice that r (t) =, t, 3t and r (s) = 0, cos(s), sin(s). The point (,, ) corresponds to t = and s = π/4. Therefore, crossing r () and r (π/4) gives a normal vector to the plane, and in this case gives 5,,. Therefore the plane is 5(x ) + (y ) + (z ) = 0. Exercise 5 (Linear Approx/Chain Rule). The resistance of a circuit consisting of three resistors connected in parallel satisfies R = R + R + R 3, where R, R, R 3 are the resistances of each of the individual resistors. (a) If R = R = 00Ω and are increasing at Ω/s and R 3 = 00Ω decreasing at Ω/s, at this moment, how fast is R changing? 3
4 (b) R, R, R 3 are measured to be 5, 40, and 50 Ohms, respectively, with a possible error of percent in each case. Estimate the maximum possible error in calculated value of R. Solution 5. (a) Take d/dt of both sides implicitly to get dr R dt = dr dr dr R dt R dt R dt. If R = R = 00 and R 3 = 00, then R = 40. Now you know everything except dr so plug in: ( dr dt = ) (b) Now use differentials: R dr = dr R dr R dr R3 3. dt, If the error is at most percent, then the maximum errors in each case is,, and, 4 5 respectively, which are your dr, dr, and dr 3. Again, plug in to get ( dr = R ). With the given values of R, R, R 3, R = 00. Plug this in to get dr, which is your 7 answer. Exercise 6 (Gradient). Find a vector tangent to x y = 8 at (, 3). Solution 6. Interpret this curve as a level curve to f(x, y) = x y at c = 8. The gradient is normal to the level curve: f(, 3) = yx y, x y ln(x) (,3) =, 8 ln(). A vector tangent to this curve is a vector orthogonal to the gradient, say 8 ln(),. Exercise 7 (Directional Derivative). Find the directional derivative of f(x, y) = xy at P (, 8) in the direction of Q(5, 4). Solution 7. First, f =, so f(, 8) =, 4. The vector P Q = 3, 4, so y xy, x xy a unit vector in this direction is u = 3 5, 4 5. Therefore, we just need f(, 8) u = = 5. Exercise 8. Find the maximum rate of change of f(x, y) at the given point and the direction (as a unit vector) in which it occurs. 4
5 (a) f(x, y) = sin(xy), at (, 0) (b) f(x, y) = 4y x at (4, ) Solution 8. The maximum rate of change is f and the direction is in the direction of the gradient. (a) in the direction 0,. (b) 65, in the direction 8 65, 65. Exercise 9. Let f(x, y, z) = x y + xyz, v = 3, 4,. (a) Calculate D v f(,, ) (b) Let r(t) be a differentiable function giving the position of a moving particle at time t such that at t = 0 the object is at the point (,, ) moving in the direction v at speed. Compute r (0). (c) If g(t) is the value of the function f at the objects position at time t, find g (0). Solution 9. (a) This is straight computation: ˆv = 3 meaning f(,, ) = 0, 5,. Therefore, 4, 3 3 D v f(,, ) = f(,, ) ˆv = , f(x, y, z) = x + yz, y + xz, xy, (b) r (0) is the tangent vector to r(t) at t = 0. The tangent vector points in the direction of motion, which you are given to be v. The speed is, so you need the magnitude to be. Therefore you are finding a vector of length in the direction of v, which is ˆv. So r (0) = ˆv, which we found in (a). (c) g(t) = f(r(t)), so g (t) = f(r(t)) r (t), and g (0) = f(r(0)) r (0). But r (0) = ˆv and since r(0) = (,, ), f(r(0)) = 0, 5,, since we found this in (a). Therefore g (0) = 0, 5, ˆv = 4 3. Notice you ended up doing the same work as in (a). Try to connect the two ideas. Exercise 0. Consider the surfaces z = x + y and x + y =. A particle moves along the itnersection of these surfaces from the point (, 0, 4) to (0,, 4) in such a way that x = t. At the time the position is (,, ), find: v, the speed, a, and decompose the acceleration into the tangential and centripetal components. Solution 0. So first we need the full r(t). We know x = t, and since x + y =, we know y = x = ( t) = t. Similarly, since z = x + y, we get z = ( t) + t = t 4t
6 Therefore r(t) = t, t, t 4t + 4. Notice that the point (,, ) corresponds to t =. So v = r (). Since r (t) =,, 4t 4, we get r () =,, 0. The speed is v =. For acceleration we need r (t): r (t) = 0, 0, 4, so r () = a = 0, 0, 4. To decompose it: the unit tangent vector is gotten by normalizing r (0): T =,, 0. Then a T = a T = 0. We can get a N using a N = a = a T T + a N N, and we know everything but N, we can solve for N: N = a a T T a N. a a T = 4. Finally, since we have Exercise. Use differentials to approximate the amount of tin in a closed (cylindrical) tin can with diameter 8 cm and height cm if the tin is.04 cm thick. Solution. You have to interpret this the right way. I can consider a cylinder with radius 4 cm (notice that you were given the diameter, not the radius) and height and compute its volume. Adding thickness decreases the volume by a certain amount, and this amount is the amount of tin that I m adding. So the amount of tin being used is approximately dv. Since V = πr h, dv = (πrh)dr + (πr )dh. You know r = 4, h =. If the thickness is.04, then the radius decreases by.04 and the height decreases by.08 (.04 each for the top and bottom). Therefore dr =.04, dh =.08. Plug these in to get dv : dv = (π 4 ).04 + π This is a negative quantity because the volume decreased. The amount of tin is just the absolute value of this. Exercise. A skier is on a mountain with height function z = f(x, y) = 00.4x.3y, where z is the height. (a) The skier is at (,, f(, )) and wants to ski downhill along the steepest path. In which direction should the skier go? Give answer as a unit vector. (b) The skier begins skiing in the direction you found in (a), so the skier heads in a direction in 3D space given by a, b, c. Find this vector. Solution. (a) The path of steepest descent happens in the direction of f. Here, f(x, y) =.8x,.6y, so f(, ) =.8,.6 =.8,.6. This vector is already a unit vector. 6
7 (b) From (a), you know the x and y components are.8 and.6, respectively. But you need c. Notice that the rate of steepest descent is f =. Intepret this as a slope: if I move unit in the direction of.8,.6, my zvalue changes by. unit in that direction means you ve moved a total of.8,.6 =, and your vertical distance changed is exactly c. So c =, so c =. Thus, your vector is.8,.6,. Exercise 3. If g(x, y) = y x e t dt, find the linear approximation to (0, 0) and use it to approximate.. e t dt. Solution 3. By Fundamental Theorem of Calculus, g x (x, y) = e x, and g y (x, y) = e y. So g x (0, 0) = and g y (0, 0) =. Since g(0, 0) = 0 0 e t dt = 0, the point is (0, 0, 0) and the normal vector is,,. Thus, the plane is x + y z = 0 = z = y x. To approximate the given integral, we notice that it corresponds to x =. and y =., so plug this into the tangent plane to get that the approximate value is. (.) =.3. Exercise 4. (a) Find a parametrization of the line through A(,, 3) and B(0,, ) such that the lines passes through A at t = and through B at t =. (b) Find where the line in (a) intersects the sphere of radius 8 centered at the origin. (c) Find an arc length parametrization of the line. Solution 4. (a) Your direction vector is AB =, 0,, so a line is,, 3 +t, 0,. The problem is that point A corresponds to t = 0 and point B corresponds to t =. To remedy this, you can shift by :,, 3 + (t ), 0,. You can check now that this goes through A at t = and B at t =. (b) The line is x = t, y = and z = 4 t. The equation of the sphere is x +y +z = 8. Plug these x, y and z into the sphere equation and solve for t. You should get t = or t = 4, so there are two points: t = corrsponds to (0,, ) and t = 4 corresponds to (,, 0). (c) You can simplify the line to,, 4 + t, 0,. The arc length parametrization of the line is gotten by essentially normalizing the direction vector: r(s) =,, 4 + s, 0,. Or you could do all the steps we did for the last midterm to get this as well. 7
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