ES.182A Problem Section 11, Fall 2018 Solutions

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1 Problem (a) z = (b) z = 2 2 ES.182A Problem Section 11, Fall 2018 Solutions Sketch the following quadratic surfaces. answer: The figure for part (a) (on the left) shows the z trace with = 0 and the z trace with = 0. It also shows two contours at z = 1 and z = 2. The figure for part (b) (on the right) shows the z trace with = 0 and a number of z traces with different values of. It also shows the z traces with = ±1. z z Problem (a) z = e (2 + 2 ) Draw level curves for each of the following. answer: The level curves are z = e (2 + 2) = constant. This is the same as = constant. More carefull: z = e (2 + 2) = c = ln(c). Since , we onl get a level curve if ln(c) 0, i.e. 0 < c 1. So the level curve z = e 1 is = 1, i.e. the circle of radius 1. The level curve z = e 4 is = 4, i.e. the circle of radius 2. The level curve z = 1 is = 0, i.e. the point at the origin. 1

2 ES.182A Problem Section 11, Fall 2018 Solutions 2 z = e 9 z = e 4 z = e (b) z = e 2 2. (For this one, ou need to remember about hperbolas.) answer: Similar to part (a), the level curves are 2 2 = constant, i.e. z = e 2 2 = e c 2 2 = c. These are hperbolas oriented with the coordinate aes. In this case, the level curves 2 2 = 0 (i.e., z = 1) are two intersecting lines = and =. z = e 9 z = e 4 z = 1 z = e 1 z = e 4 z = e z = 1 Problem (a) Find the equation of the tangent plane to z = 2 at the point (1,1,1). answer: f = 2, f = 2 f (1, 1) = 1, f (1, 1) = 2 tangent plane is z 1 = ( 1) + 2( 1). (b) Use our answer in part (a) to estimate the value of z at (, ) = (1.1, 0.8). answer: Part (a) gives the tangent plane approimation formula z In our case, we have = 0.1, = 0.2. So, z = Thus, z = 1 + z 0.7.

3 ES.182A Problem Section 11, Fall 2018 Solutions 3 Problem (a) F(, ) =, = i j. Sketch the following vector fields. answer: This is called a tangential vector field because all of the vectors are tangential to circles centered at the origin. The arrows point in a clockwise direction. As the circles get bigger the vectors get longer. (b) F(, ) = /r, /r = ( /r) i+( /r) j. Here r is the radial distance, r = answer: This is called a radial vector field because all of the vectors point along the line from the point to the origin. In this case. the point inwards towards the origin. Each vector is a unit vector

4 ES.182A Problem Section 11, Fall 2018 Solutions 4 Problem (a) f(, ) = Compute the gradients of the following functions. answer: f(, ) = , (a) f(, ) = e answer: f(, ) = e, e Problem Use the following level curves to estimate the asked for directional derivatives as z. The horizontal and vertical scales are the same. s z = 2 B E2 D1 F1 z = 9 E1 z = 7 z = 8 A z = 3 z = 4 z = 5 z = 6 F2 C F3 D2 z = 5 z = 4 z = 3 z = (a) Let û be the vector shown starting at the point A. Estimate dz ds answer: We use the scale shown to measure from A to the level curve z = 7 going along û. It is about s = 0.6 In going between the level curves we have z = 1. Thus, dz ds z A,û s (b) Draw in the gradient at the point B. answer: The gradient is perpendicular to the level curves and points in the direction of increase. The vector shown starting at B is in that direction. The length of the gradient is the directional derivative in the same direction. We estimated that b computing z s 1 = 2. (The Deltas are between B and the level curve z = It is not a good idea to estimate (c) Estimate answer: derivative in the and at A. and A,û in order to estimate the gradient. is the directional derivative in the î direction. Likewise, is the direction ĵ direction. We estimated these b measuring Deltas between A and the

5 ES.182A Problem Section 11, Fall 2018 Solutions 5 level curve z = 7. 1 A 0.8 = 1.2, 1 A Note: We d get ver different answers if we measure Deltas between A and the level curve z = 9. In fact, in the ĵ we wouldn t ever reach the curve z = 9. The level curves are not fine enough to get reall good estimates. (d) Mark all the points on the z = 9 curve where = 0. answer: The partial with respect to is 0 when the level curve has a horizontal tangent, i.e. there is no change in the î direction. Equivalentl, the partial with respect to is 0 when the gradient is in the ĵ direction. This happens at the points D1 and D2 marked on the graph. (e) Mark all the points on the z = 7 curve where = 0. answer: This is similar to part (d). The partial with respect to is 0 when the level curve has a vertical tangent. These points are labeled E1 and E2 on the graph. (f) Mark all the critical points on the figure. Sa whether the are maima, minima or neither. answer: The critical points (gradient = 0) are at peaks, valles and mountain passes (saddles). Peaks and valles are seen at the center of level curves that are closed loops. Saddles are seen where the level curve crosses itself. We have maima at F 1 and F 3. The point F 2 is a saddle point. These are all the critical points. (g) At the point C, which direction gives the greatest rate of increase. answer: This is a trick question. The rate of increase is 0 in all directions! (h) Sketch the surface in three dimensions that has these level curves answer: NEEDED f Problem is the directional derivative in what direction? answer: The i direction. Etra problems if ou have time Problem Sketch the quadratic surface z 2 = If ou like this ou can tr the ellipsoid 2 /a /b 2 + z 2 /c 2 = 1 and the hperboloids of one ( z 2 = 1) and two ( z 2 = 1) sheets. answer: NEEDED Problem = 2.8. (a) A rectangle has sides and. Approimate the area for = 2.1 and

6 ES.182A Problem Section 11, Fall 2018 Solutions 6 answer: We have A =. Let ( 0, 0 ) = (2, 3), =.1, =.2. Since A A = and = the tangent plane approimation formula is A A + A 0 = 3(.1) + 2(.2) =.1. 0 Therefore A 6.1 = 5.9. (b) For, near ( 0, 0 ) = (2, 3) which has a greater affect on the area: a change in or an equal change in? answer: A change in because A = 3 is greater than A = 2. Problem Make up a function (with polnomials) and compute the directional derivative in a few directions. answer: You ll need to suppl the answers. All of them will use the formula df ds = f P û. P,û Problem Give the linearization of e cos at (0, 0). (Linearization is the same thing as finding the tangent plane approimation.) answer: f(, ) = e cos f = e cos, f = e sin. f(0, 0) = 1, f (0, 0) = 1, f (0, 0) = 0 f(, ) 1 +.