CCNY 203 Fall 2011 Final Solutions

Transcription

1 CCNY Fall Final Solutions a: Take cross products of the displacement vectors to get a normal to plane P: Cross@8,, < - 8,, -<, 8,, < - 8,, <D 8-,, -< That gives an equation of the plane from the normal vector and a point: - HxL + Hy - L - Hz - L ã - x + H- + yl - H- + zl ã b: The direction of the line is the same as the normal vector above, so parameterizing the line as P + t V gives l = 8,, < + t 8-,, -< 8- t, + t, - t< If we write this in standard form, we get x = - t, y = + t, z = - t. c: We use dot product to measure the angle: 8,, -<.8-,, -< The dot product is not zero and the vectors are not scalar multiples, so the vectors are neither perpendicular nor parallel. a: We take the gradient: f = z + z LogAx + y E z + z LogAx + y E gradf = 8D@f, xd, D@f, yd, D@f, zd< : x z x + y, y z x + y, + LogAx + y E> At the desired point, we substitute to get: gradf ê. 8x Ø, y Ø, z Ø < 8,, < The directional derivative is obtained by dotting the gradient with a unit vector in the desired direction: u = 8,, 7< - 8,, < :,, > 5 5 Together@8,, <. ud 7 5 b: This the direction of the gradient: <,,>.

2 fallsols.nb c: The appropriate version of the chain rule here is E = E s x x + E s y y + E s z z s which at the relevant points gives µ + µ + H + lnhll = + ln(). Or you can substitute but that is more tedious and prone to error. : Taking the first partials and setting them to zero and solving gives: f = x - x y + 8 y x - x y + 8 y Solve@8D@f, xd ã, D@f, yd ã <D 88x Ø, y Ø <, 8x Ø, y Ø << At the first point, we evalute the discriminant to get: D@f, x, xd D@f, y, yd - D@f, x, yd ê. 88x Ø, y Ø << 8-< So there is a saddle at (,). At the second point, we evalute the discriminant to get: D@f, x, xd D@f, y, yd - D@f, x, yd ê. 88x Ø, y Ø << 8< Since it is positive, we look at f xx : D@f, x, xd ê. 88x Ø, y Ø << 86< So there is a relative minimum at (,). a: We sketch the region in the plane (it s a circle tangent to the y-axis at the origin) and convert to polar: ParametricPlotB8t Cos@qD Cos@qD, t Cos@qD Sin@qD <, :q, -p, p >, 8t,, <F

3 fallsols.nb p -p 9 Cos@qD r r r q b: We differentiate and evaluate at the point and get: f = x Iy + M D@f, xd D@f, yd D@f, xd ê. 8x Ø, y Ø -< D@f, yd ê. 8x Ø, y Ø -< x I + y M + y x y -6 which gives an equation of the tangent plane as: z - 9 ã Hx - L - 6 Hy + L 5: For the volume via a double integral, taking the integral of the (top surface - bottom surface) over the shadow to get: 79 x HHx + yl - xl y x

4 fallsols.nb ParametricPlotDA88x, y, x<, 8x, y, x + y<<, 8x,, <, 9y, x, =E 5: Alternatively, volume via a triple integral: 79 x x+y z y x x 6a: Divergent by comparison with the harmonic series. Limit comparison does not work as the limit of the ratio does not exist. 6b: Convergent by alternating series test, and the absolute values are divergent by comparison with the harmonic series, so the series is conditionally convergent. 6c: Absolutely convergent by the ratio test. 7: We use the ratio test to get convergence on the interval -9<x<. For x=, divergent by comparison with the harmonic series. For x=-9, converent by the alternating series test. So the power series convergenes on the interval [-9,). 8a: We use spherical coordinates to find the center of mass, and the fact that the relevant part of the cone has equation f = p : p mass = p p p r r Sin@fD r f q

5 fallsols.nb 5 zbar = mass 5 p p p r r Cos@fD r Sin@fD r f q ShowB:ParametricPlotDB8 Cos@qD Sin@fD, Sin@qD Sin@fD, Cos@fD <, 8q,, p<, :f, p, p >F, ParametricPlotDB : t Cos@qD SinB p F, t Sin@qD SinB p F, t CosB p F >, 8q,, p<, 8t,, <F>F 8b: We consider horizontal and vertical paths approaching the origin, which have limits of, x = t; y = ; LimitB x Sin@yD, 8t Ø <F x + y 8< x = ; y = t; LimitB x Sin@yD, 8t Ø <F x + y 8< Those were the same, so either we should try to prove that limit exists or we should look at another direction. We consider coming in along the line y=x, which gives a limit of using that the limit of sinhxl is as x approaches, so x therefore the limit does not exist:

6 6 fallsols.nb x = t; y = t; LimitB x Sin@yD, 8t Ø <F x + y 8< Converting to polar coordinates is another approach- we can see that there are different limits for different values of q as r approaches zero, which means the limit cannot exist: x = r Cos@qD; y = r Sin@qD; SimplifyB x Sin@yD F x + y Cos@qD Sin@r Sin@qDD r Cos@qD Sin@r Sin@qDD LimitB r 8< Cos@qD Sin@r Sin@qDD LimitB r 8< ê. :q Ø p >, 8r Ø <F ê. 8q Ø <, 8r Ø <F 9a: we take the nearby point of (,) and use differentials from there: f = x-y Cos@x - D x-y Cos@ - xd f ê. 8x Ø, y Ø < D@f, xd ê. 8x Ø, y Ø < D@f, yd ê. 8x Ø, y Ø < - which gives us the approximation of f at that point as +(-.) + -(.) =.9 9b: we take the gradient of the function since the surface is given implicily: f = Sin@x - yd + y -z y -z + Sin@x - yd f ê. 8x Ø, y Ø, z Ø < gradf = 8D@f, xd, D@f, yd, D@f, zd< ê. 8x Ø, y Ø, z Ø < 8, 6, -< which gives an equation of Hx - L + 6 Hy - L - Hz - L = for the tangent plane there. a: For surface area, we integrate + f x + f y over the shadow down on the x y plane, then switch to polar, to get

7 fallsols.nb 7 p J- + N p 6 + r r r q b: this is easier to recognize when we complete the square for x to get into the form Hx - L - y - z =, where we can regonize this as a hyperboloid of sheets opening up along the +x-axis and -x axis The vertices are (,,) and (,,). a: this is a sideways heart-shaped region, and we set up a mass integral in cylindrical coordinates PolarPlot@ + Cos@tD, 8t,, p<d - - p mass = p + Cos@qD z r z r q +H-L : the correct series is n n= n! x n. To approximate f I M, we take the first few terms as + I M I M + I -! M I M + I! M I M - I 7! M I M +... By the alternating series test, the series converges and the error is no more than the first omitted term. So since the second term is, the first term is not guarranteed to be enough. But 8 since the third term is -, the error in using the first two terms is less than the desired.. So the approximation is -. 8 For the series of the derivative, since the series converges absolutely, we can differentiate each term to get n= n +H-Ln x n- as a description of the series for the derivative, which could be re-indexed if desired. The first n! three non-zero terms are:

8 8 fallsols.nb n + H-Ln n! n= x n x - 7 x