Math 2A Vector Calculus Chapter 11 Test Fall 07 Name Show your work. Don t use a calculator. Write responses on separate paper.

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1 Math A Vector Calculus Chapter Test Fall 7 Name Show our work. Don t use a calculator. Write responses on separate paper.. Consider the nice, smooth unction z, whose contour map is shown at right. a. Estimate unction alues to the nearest tenth to ill in the blank cells in the table below: \ b. Use the alue in our table aboe to estimate (.4,.) and (.4,.) to the nearest tenth. c. Let be the ector rom P(.6,.4) to Q(.,.5) Compute D.4,. in two was: z as h and as z d. Let be the ector rom P(.,.4) to Q(.6,.5) Compute D.4,. in two was: as z h and as z.. B considering dierent lines o approach, show that the unction has no limit as (, ) (,)., 3. Use polar coordinates to proe that the limit eists: e lim 4. Sketch leel cures (,) =, (,) = ½ and (,) = ½ or the i,, unction,. i,,

2 5. Find points on the surace + z + z z = where the tangent plane is parallel to the -plane. 6. Find an equation or the plane tangent to the parametricall deined surace,, cos,cos sin,cos z u u where (u,) = (,π/). 7. Show that, arctan / satisies the two dimensional Laplace equation,. 8. Find the direction in which increases and decreases most rapidl at P and the rates at which changes in these directions. a. (,) = + cos, P (,). b. (,,z) = z ln( + ), P (,,) 9. Consider 3 3, 3 a. Find the critical points. b. Find all maima, minima and saddle points and ealuate the unction at those points.. A lat circular plate has the shape o the region +. The plate, including the boundar where + =, is heated so the temperature at an point (, ) is T,. Find the absolute ma. and min. alues o (, ) = on the ellipse + 4 = 8 in two was: a. B using the parameterization, cos t, sin t b. B using Lagrange multipliers.. Find a leel surace or the densit unction the tangent plane + 3 z = 3.,, z z that has

3 Math A Vector Calculus Chapter Test Solutions Fall 9. Consider the nice, smooth unction z, whose contour map is shown at right. a) Estimate unction alues to the nearest tenth to ill in the blank cells in the table below: \ b) Use the alue in our table aboe to estimate (.4,.) and (.4,.) to the nearest tenth. z.3..4,..6. z.3..4,..5.4 c) Let be the ector rom P(.6,.4) to Q(.,.5) Compute D.4,. in two was: as z h and as z SOLN: PQ..6, z So, D.4,. or h h.4,..4,... z,.5, d) Let be the ector rom P(.,.4) to Q(.6,.5) Compute D.4,. in two was: as z h and as z. SOLN: PQ.6., z.37 So, D.4,..3 or h.7.4,..4,..4 z,.5, B considering dierent lines o approach, show that the unction has no limit as (, ) (,).,

4 SOLN: Along the line =, lim lim lim,, lim lim lim lim does not eist because 3. Use polar coordinates to proe that the limit eists: e lim r e e SOLN: Since goes to ininit onl i r goes to ininit, lim lim r r Now it ma happen that r goes to ininit but, =, but this wouldn t appl to this limit. 4. Sketch leel cures (,) =, (,) = ½ and (,) = ½ or the i,, unction,. i,, SOLN: This shows, b the wa, that there is no limiting alue o z as (,) approaches (,). 5. Find points on the surace + z + z z = where the tangent plane is parallel to the -plane. SOLN: We need z and z. Each o these equations describes a plane, and the intersection o these planes is the line r t t, t, t,, t,,.

5 So since ( + z) = the intersection o the line with the leel surace occurs where z z = or t + t t = t(t ) = so either t = or t = ½ whence the points where tangent plane is horizontal are (,,) and ( ½, ½, ½). To help isualize what s going on here, ou might sole the equation or the surace or z: 4 z z z. We can isualize this in Mathematica Plot3D[{{( Sqrt[( ) 4 ( )]) / }, with the ollowing command: {( Sqrt[( ) 4 ( )]) / }, {},{/ }},{, 5,5},{, 5,5}] The graph shows a tilted hperboloid o one sheet. 6. Find an equation or the plane tangent to the parametricall deined surace,, cos,cos sin,cos z u u where (u,) = (,π/). SOLN: The point o tangenc is cos /,cos sin /,cos,, The tangent ector r ', / sin /,cos cos /,,, but the other one is r u ', /, sin sin /, sin, so what to do? What does this mean? It could mean there s a cusp in that direction, but this doesn t mean there isn t a tangent plane. It could be we could do a directional deriatie in some other direction and get another ector or the cross product normal to the tangent plane, but the simplicit o the ormula or the surace suggests we tr to eliminate the parameters and get a rectangular orm or the equation o the surace. Obsering that z = cos(u) so = z sin() leads to z sin z cos z so the solution set to this rectangular equation can be iewed rom a higher dimension as the leel surace w = or the densit unction w,, z z. Thus a normal to the tangent plane at (,,) can be ound b ealuating the gradient ector there: w z z so an equation or the,,,,,, tangent plane is obtained b requiring that the normal be perpendicular to an arbitrar ector in the plane:,,,, z z z To isualize this in Mathematica, the ollowing command will graph the surace and tangent plane:

6 ParametricPlot3 D[{{ Cos[ ], Cos[ u] Sin[ ], Cos[ u]},{ / Pi, u / Pi, u / Pi}},{ u, Pi, Pi},{, Pi, Pi }] It s worth eamining this igure in detail. For instance, u = is a circle o radius in the plane z =. 7. Show that, arctan / satisies the two dimensional Laplace equation, SOLN:. d d, arctan /, arctan /,, / / d d, 8. Find the direction in which increases and decreases most rapidl at P and the rates at which changes in these directions. a) (,) = + cos, P (,). SOLN:, sin, sin, so grows at a rate o / in that direction., b) (,,z) = z ln( + ), P (,,)

7 SOLN: 9. Consider z z,,,,ln,, So is growing at a rate o / is that direction. 3 3, 3 a) Find the critical points. SOLN: 3 3 and 4,, 3 3, substituting, we get or, leading to critical points (,) and (, ). b) Find all maima, minima and saddle points and ealuate the unction at those points. SOLN: D is positie at (, ) where = 6 <, so this is a local maimum. The point (,) has D < so the point is a saddle. In the diagram below, the local ma at (-,-,) seems eident. The saddle is a little more subtle. 3 Look at the cures r t, t,t 3t and r t, t, 3t are shown on the plot and ou can see the irst is curing up at (,,) and the other is curing down A lat circular plate has the shape o the region +. The plate, including the boundar where + =, is heated so the temperature at an point (, ) is T,. Find the etreme temperatures o the plate and where the occur. SOLN: The critical points are where T and T 4. So there s a critical point at (½, ) where = indicates that (½,, -¼) is a local (turns out to be global) minimum. On the boundar we

8 hae the cure r cos t,sin t, cos t sin t cost cos t,sin t, sin t cost cos t,sin t, cost cos t the tangent line is horizontal when dz dt sin t cos t sin t sin t cos t t k or t dz. 3 We eamine the point where t = and ind (,,) where cost cos t 3 and so that s a local dt min on the edge, (ront right o edge in the image below,) but neither a ma nor a min. Where t = π, d z (,,) has, so that s also a local min along the edge, but neither a local min nor a local ma on dt the surace. At t = π/3 the point (-½, 3/,9/4) and at t = π/3 the point ( ½, 3/,9/4) hae d z, so those are global maima. dt. Find the absolute ma. and min. alues o (, ) = on the ellipse + 4 = 8 in two was a) B using the parameterization, cos t, sin t SOLN: Along the path, cos t, sin t, z = 4cos t sin t = sint so z = 4cost = i t = an odd multiple o π/4. At π/4 and 5π/4 z = 4 so (,, ) and (,,) are global maima and at 3π/4 and 7π/4 z = 4 so (,, ) and (,, ) are global minima. b) B using Lagrange multipliers. SOLN: g,,8 leads to the sstem Substituting rom the second to the irst, 6, we know that either 8 = or λ = ±¼. I = then = and then constraint 4 8 can t 4 8 be met so λ = ±¼ which means = ±/ and substituting into the the ellipse equation, 8 meaning that Ater inestigating we determine that the global ma is occurring at (,) and (, ) and the global min is occurring at (,) and (, ).

9 . Find a leel surace or the densit unction,, z z that has the tangent plane + 3 z = 3. SOLN: The normal to the leel surace will be parallel to the normal to the plane i,, z,3, so that λ = = /3 = z and substituting into the equation o the plane, λ + 9λ/ λ/ = 6λ = 3 or λ = / and thus /,3 / 4,/ 4 / 3 / 4 / 4 3 / 4. So the leel surace is z 3/ 4

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