Quiz 6 Practice Problems

Quiz 6 Practice Problems Practice problems are similar, both in difficulty and in scope, to the type of problems you will see on the quiz. Problems marked with a are for your entertainment and are not essential. EXPECTED SKILLS: Be able to compute an equation of the tangent plane at a point on the surface z = f(x, y). Given an implicitly defined level surface F (x, y, z) = k, be able to compute an equation of the tangent plane at a point on the surface. Know how to compute the parametric equations (or vector equation) for the normal line. Be able to use gradients to find tangent lines to the intersection curve of two surfaces. Be able to find (acute) angles between tangent planes and other planes. Be able to compute the local linear approximation for a function of two or more variables at a given point. Be able to use a local linear approximation to estimate a given quantity. Be able to use local linearity to estimate the change in the function. Be able to bound the error in approximating a function by way of local linearity. 1

PRACTICE PROBLEMS: For problems 1-4, find two unit vectors which are normal to the given surface S at the specified point P. 1. S : x y + z = 7; P ( 1,, ). S : x y + z = 11; P ( 1,, ). S : z = y 4 ; P (, 1, 1) 4. S : z = x cos(xy); P ( 1, π ), For problems 5-9, compute equations of the tangent plane and the normal line to the given surface at the indicated point. 5. S : ln(x + y + z) = ; P ( 1, e, 1) ( ) 6. S : x + y + z = 1; P,, 7. S : z = sin 1 ( x y ) ( ; P 1,, π ) 4 8. S : x xy + z = 9; P (,, ) ( π 9. S : z = x cos(x + y); P, π ) π, 4 10. Consider the surface S 1 : x + y = 5 and S : z = x (a) Find an equation of the tangent line to the curve of intersection of S 1 and S at the point (, 4, 1). (b) Find the acute angle between the planes which are tangent to the surface S 1 and S at the point (, 4, 1).

11. Consider the surface S 1 : z = x y and S : y + z = 10 (a) Find an equation of the tangent line to the curve of intersection of S 1 and S at the point (, 1, ). (b) Find the acute angle between the planes which are tangent to the surface S 1 and S at the point (, 1, ). 1. Find all points on the ellipsoid x + y + z = 7 where the tangent plane is parallel to the plane 4x + 4y + 1z =. 1. Find all points on the hyperboloid of one sheet x + y z = 9 where the normal line is parallel to the line which contains points A(1,, ) and B(7, 6, 5). 14. Show that every plane which is tangent to the cone z = x + y must pass through the origin. (HINT: Compute the equation of the plane which is tangent to the surface at the point P 0 (x 0, y 0, z 0 ) and see what happens.) For problems 15-19, find the local linear approximation L(x, y) of the given function at the specified point. 15. f(x, y) = x y ; P (1, ) 16. f(x, y) = x + y ; P (, 1) x y ( 17. f(x, y) = e x sin y; P ln, π ) 18. f(x, y) = ln(x y ); P (, ) ( ) x 19. f(x, y) = tan 1 ; P (1, 1) y

0. Find the local linear approximation of the function f(x, y) = x y at (1, ), and use it to approximate f(0.98,.01). 1. Suppose that f is a differentiable function at the point (, ) with f(, ) = 1, f x (, ) = 5, and f y (, ) =. Estimate f(1.98,.01).. Find the local linear approximation L(x, y, z) to f(x, y, z) = x y + xz at the point P ( 1,, 1).. Verify that e x cos y 1 + x for (x, y) near (0, 0). 4. Verify that (x + y) 1x + 1y 16 for (x, y) near (1, 1). 5. At a particular point P (x 0, y 0 ), the local linear approximation of f(x, y) = xy + y is L(x, y) = 15 + x + 8y. What is the point P? 6. Let z = f(x, y) = x +xy y. If x changes from to.05 and y changes from to.96, approximate the value of z = f(.05,.96) f(, ) by way of local linearity. 7. The base radius and height of a right circular cone are measured as 10 cm and 5 cm, respectively, with a possible error in measurement of as much as 0.1 cm in each. Use local linearity to estimate the maximum relative error and percentage error in the calculated volume of the cone. (a) Newton s method for approximating a root of an equation f(x) = 0 can be adapted to approximating a solution of a system of equations f(x, y) = 0 and g(x, y) = 0. The surfaces z = f(x, y) and z = g(x, y) intersect in a curve that intersects the xy-plane at the point (r, s), which is the solution of the system. If an initial approximation (x 1, y 1 ) is close to this point, then the tangent plane to the surfaces at (x 1, y 1 ) intersect in a straight line that intersects the xy-plane in a point (x, y ), which should be closer to (r, s). Show that 4

x = x 1 fg y f y g f x g y f y g x and y = y 1 f xg fg x f x g y f y g x where f, g, and their partial derivatives are evaluated at (x 1, y 1 ). If we continue this procedure, we obtain successive approximations (x n, y n ). (b) It was Thomas Simpson (1710-1761) who formulated Newton s method as we know it today and who extended it to functions of two variables as in part (a). The example that he gave to illustrate the method was to solve the system of equations x x + y y = 1000 x y + y x = 100. In other words, he found the points of intersection of the two curves above. Use the method of part (a) to find the coordinates of the points of intersection correct to six decimal places. 5

SOLUTIONS 1. Set f(x, y, z) = x y + z. Then f(x, y, z) =, 1, 1 so that f( 1,, ) =, 1, 1 is normal to the level surface S at P. Normalize f( 1,, ) to get the desired unit vectors:, 1, 1 ± + ( 1) + 1 = ± 6, 1 1,. 6 6. Set f(x, y, z) = x y + z. Then f(x, y, z) = x,, z so that f( 1,, ) =,, 4 is normal to the level surface S at P. Normalize f( 1,, ) to get the desired unit vectors:,, 4 ± ( ) + ( ) + 4 = ±, 4,. 9 9 9. Set f(x, y, z) = y 4 z. Then f(x, y, z) = 0, 4y, 1 so that f(, 1, 1) = 0, 4, 1 is normal to the level surface S at P. Normalize f(, 1, 1) to get the desired unit vectors: 0, 4, 1 ± 0 + ( 4) + ( 1) = ± 0, 4, 1. 17 17 4. Set f(x, y, z) = x cos(xy) z. Then f(x, y, z) = x cos(xy) + x y sin(xy), x sin(xy), 1 so that f ( 1, π ) (, = cos π ) + π ( sin π ) (, sin π ), 1 π, 1, 1 is normal to the level surface S at P. Normalize f ( 1, π ), to get the desired unit vectors: π/, 1, 1 ± ( π/) + ( 1) + ( 1) = ± π π /4 +, 1 π /4 +, 1. π /4 + 6

5. Set f(x, y, z) = ln(x + y + z). Then f(x, y, z) = 1/(x + y + z), 1/(x + y + z), 1/(x + y + z) so that f ( 1, e, 1) = 1/e, 1/e, 1/e is normal to the level surface S at P. The equation of the tangent plane is thus e (x + 1) + e (y e ) + e (z 1) = 0 x + y + z = e and the normal line is l(t) = 1, e, 1 + t 1, 1, 1. [NOTE: the gradient is parallel to 1, 1, 1, and so we can use this vector instead in the equation of the line for the sake of simplicity.] 6. Set f(x, y, z) = x + y + z. Then f(x, y, z) = x, y, z so that f( /, /, /) = /, /, / is normal ( to the level surface S at P. The equation of the tangent plane is thus ) x + ( ) ( y + ) z = 0 x + y + z = and the normal line is l(t) = /, /, / + t 1, 1, 1. [NOTE: this is the same thing that we did in the previous problem.] 7. Set f(x, y, z) = sin 1 (x/y) z. Then f(x, y, z) = 1/(y 1 (x/y) ), x/(y 1 (x/y) ), 1 so that f( 1,, π/4) = 1, /, 1 is normal to the level surface S at P. The equation of the tangent plane is thus (x + 1) + (y + ( ) z π ) = 0 x + 4 y z = π and the normal line is 4 l(t) = 1,, π/4 + t 1, /, 1. 8. Set f(x, y, z) = x xy + z. Then f(x, y, z) = x y, x, z so that f(,, ) =,, 6 is normal to the level surface S at P. The equation of the tangent plane is thus (x ) (y ) + 6(z ) = 0 x y + z = 9 and the normal line is l(t) =,, + t 1, 1,. 9. Set f(x, y, z) = x cos(x + y) z. Then f(x, y, z) = cos(x + y) x sin(x+y), x sin(x+y), 1 so that f(π/, π/, π/4) = cos(π/+ π/) (π/) sin(π/ + π/), (π/) sin(π/ + π/), 1 = cos(5π/6) 7

(π/) sin(5π/6), (π/) sin(5π/6), 1 = / π/4, π/4, 1 is normal to ( the level surface S at P. The equation of the tangent plane is ) ( thus π ( x π ) π ( y π ) ) π z + = 0 and the 4 4 4 normal line is l(t) = π/, π/, π /4 + t π +, π, 4. 10. (a) Set f(x, y, z) = x +y and g(x, y, z) = x z. Then f(x, y, z) = x, y, 0 and g(x, y, z) = 1, 0, 1 so that f(, 4, 1) = 6, 8, 0 and g(, 4, 1) = 1, 0, 1 are both orthogonal to the curve of intersection of S 1 and S at (, 4, 1). Therefore the tangent line to the curve of intersection is parallel to v = f(, 4, 1) g(, 4, 1) : v = 6, 8, 0 1, 0, 1 i j k = 6 8 0 1 0 1 8 0 = 0 1 i 6 0 1 1 j + 6 8 1 0 k = 8i + 6j + 8k. Since the tangent line is parallel to 8, 6, 8 = 4,, 4 and goes through the point (, 4, 1), it follows that the tangent has vector equation, 4, 1 + t 4,, 4. (b) The acute angle between the tangent planes to S 1 and S is the same as the acute angle between their normals, which are consequently f(, 4, 1) and g(, 4, 1). The angle between f(, 4, 1) and g(, 4, 1) is 8

( ) ( f(, 4, 1) cos 1 g(, 4, 1) = cos 1 f(, 4, 1) g(, 4, 1) ) (6)( 1) + (8)(0) + (0)( 1) 6 + 8 + 0 ( 1) + 0 + ( 1) ( ) ( 6 = cos 1 10 = cos 1 ) 5. Because f(, 4, 1) g(, 4, 1) < 0, the ( above ) angle is obtuse. To obtain an acute angle, take π cos 1 5. 11. (a) Set f(x, y, z) = x y z and g(x, y, z) = y + z. Then f(x, y, z) = x, y, 1 and g(x, y, z) = 0, y, z so that f(, 1, ) = 4,, 1 and g(, 1, ) = 0,, 6 are both orthogonal to the curve of intersection of S 1 and S at (, 1, ). Therefore the tangent line to the curve of intersection is parallel to v = f(, 1, ) g(, 1, ) : v = 4,, 1 0,, 6 i j k = 4 1 0 6 1 = 6 i 4 1 0 6 j + 4 0 k = 10i 4j + 8k. Since the tangent line is parallel to 10, 4, 8 = 5, 1, 4 and goes through the point (, 1, ), it follows that the tangent has vector equation, 1, + t 5, 1, 4. (b) The acute angle between the tangent planes to S 1 and S is the same as the acute angle between their normals, which are conse- 9

quently f(, 1, ) and g(, 1, ). The angle between f(, 1, ) and g(, 1, ) is ( ) ( f(, 1, ) cos 1 g(, 1, ) = cos 1 f(, 1, ) g(, 1, ) ) (4)(0) + ( )() + ( 1)(6) 4 + ( ) + ( 1) 0 + + 6 ( ) 10 = cos 1. 1 40 Because f(1, ) g(, 1, ) < 0, the ( above angle ) is obtuse. To 10 obtain an acute angle, take π cos 1. 1 40 1. Set f(x, y, z) = x + y + z. Then f(x, y, z) = x, 4y, 6z. Keeping in mind that f(x, y, z) is normal to the tangent plane of the ellipsoid at the point (x, y, z), 4, 4, is normal to the plane 4x + 4y + 1z =, and two planes are parallel precisely when their normals are parallel, we want to find all points (x, y, z) such that there is a scalar c with x, 4y, 6z = c 4, 4, 1. Setting corresponding components equal, we have x = 4c, 4y = 4c, and 6z = 1c. So we need x = c, y = c, and z = c subject to the constraint x + y + z = 7, as we are looking for a point on the ellipsoid. Hence (c) +c +(c) = 7 4c +c +1c = 18c = 7 c = 4 c = ±. If c =, then we get the point (4,, 4), and if c =, then we get the point ( 4,, 4). 1. Set f(x, y, z) = x + y z. Then f(x, y, z) = x, y, z. Keeping in mind that f(x, y, z) is parallel to the normal line to the ellipsoid 10

at the point (x, y, z) and AB = 7 1, 6, 5 = 6, 4,, we want to find all points (x, y, z) such that there is a scalar c with x, y, z = c 6, 4,. Setting corresponding components equal, we have x = 6c, y = 4c, and z = c. So we need x = c, y = c, and z = c subject to the constraint x + y z = 9, as we are looking for a point on the hyperboloid. Hence (c) +(c) ( c) = 9 9c +4c c = 9 1c = 9 c = 9 1 = 4 c = ± If c =, then we get the point ( /,, /), and if c =, then we get the point ( /,, /). 14. Let F (x, y, z) = x + y z. The given surface is the level surface F (x, y, z) = 0. So F (x 0, y 0, z 0 ) = x 0, y 0, z 0 is normal to the surface at the point (x 0, y 0, z 0 ). Thus, an equation of the plane which is tangent to the given surface at the point (x 0, y 0, z 0 ) is x 0 (x x 0 ) + y 0 (y y 0 ) z 0 (z z 0 ) = 0. This can be rewritten as x 0 x + y 0 y z 0 z x 0 y 0 + z 0 = 0. Now since (x 0, y 0, z 0 ) is the point of tangency, it must also be a point on the surface. Thus x 0 + y0 = z0, which implies that x 0 y0 = z0. Using this fact, the equation of the tangent plane can be rewritten as 11

x 0 x + y 0 y z 0 z x 0 y 0 + z 0 = 0 x 0 x + y 0 y z 0 z z 0 + z 0 = 0 x 0 x + y 0 y z 0 z = 0, and (0, 0, 0) satisfies this equation. Thus, since (x 0, y 0, z 0 ) was an arbitrary point on the surface and its tangent plane passes through the origin, we have that all tangent planes to the surface must pass through the origin. 15. f(1, ) = 1 =, f x (x, y) = x, f y (x, y) = y, and so L(x, y) = f(1, )+f x (1, )(x 1)+f y (1, )(y 1) = +(x 1) 4(y 1). 16. f(, 1) = + 1 1 =, f (x y) (x + y) y x(x, y) = = (x y) (x y), f y(x, y) = (x y) + (x y) x =, and so (x y) (x y) L(x, y) = f(, 1)+f x (, 1)(x )+f y (, 1)(y 1) = (x )+4(y 1). 17. f so ( ln, π ) = e ln sin π =, f x(x, y) = e x sin y, f y (x, y) = e x cos y, and L(x, y) = f ( ln, π ( )+f x ln, π ) ( (x ln )+f y ln, π ) ( y π ) = +(x ln ). 1

18. f(, ) = ln( ( ) ) = ln 1 = 0, f x (x, y) = x x y, f y(x, y) = y, and so x y L(x, y) = f(, )+f x (, )(x )+f y (, )(y ) = 4(x ) (y ). 19. f(1, 1) = tan 1 = π 4, f x(x, y) = x/y 1 + (x/y) = x, and so x + y 1/y 1 + (x/y) = y x + y, f y(x, y) = L(x, y) = f(1, 1)+f x (1, 1)(x 1)+f y (1, 1)(y 1) = π 4 +1 (x 1) 1 (y 1). 0. f(1, ) = = x 5 = 5, f x (x, y) = x y, y f y (x, y) =, and so x y L(x, y) = f(1, )+f x (1, )(x 1)+f y (1, )(y ) = 5 5 (x 1) 5 (y ). Hence f(0.98,.01) L(0.98,.01) = 5 5 ( 0.0) 5 (0.01) = 5 + 0.6(0.0) 0.4(0.01) = 5 + 0.01 0.004 = 5.08 1

1. f(1.98,.01) L(1.98,.01) = f(, ) + f x (, )(1.98 ) + f y (, )(.01 ) = 1 + 5( 0.0) (0.01) = 1 0.1 0.0 = 0.88. f( 1,, 1) = () 1 = 6, f x (x, y, z) = 6x + z, f y (x, y, z) = 4y, f z (x, y, z) = xz, and so L(x, y, z) = f( 1,, 1) + f x ( 1,, 1)(x + 1) + f y ( 1,, 1)(y ) + f z ( 1,, 1)(z 1) = 6 5(x + 1) 8(y ) (z 1) = 8 5x 8y z.. Set f(x, y) = e x cos y. Then f(0, 0) = 1, f x (x, y) = e x cos y, f y (x, y) = e x sin y, and so L(x, y) = f(0, 0)+f x (0, 0)(x 0)+f y (0, 0)(y 0) = 1+e 0 cos 0 x e 0 sin 0 y = 1+x. Hence e x cos y 1 + x for (x, y) near (0, 0). 4. Set f(x, y) = (x+y). Then f(1, 1) = (1+1) = 8, f x (x, y) = (x+y) = f y (x, y), and so L(x, y) = f(1, 1) + f x (1, 1)(x 1) + f y (1, 1)(y 1) = 8 + () (x 1) + () (y 1) = 8 + 1x 1 + 1y 1 = 1x + 1y 16. 14

Hence (x + y) 1x + 1y 16 for (x, y) near (1, 1). 5. f x (x, y) = y, f y (x, y) = x + y, and so L(x, y) = f(x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ) = x 0 y 0 + y 0 + y 0 (x x 0 ) + (x 0 + y 0 )(y y 0 ). Yet L(x, y) = 15+x+8y. Comparing coefficients shows us that y 0 = and x 0 + y 0 = 8 x 0 + 6 = 8 x 0 =. Checking for consistency, we thus have that x 0 y 0 + y 0 + y 0 (x x 0 ) + (x 0 + y 0 )(y y 0 ) = ()() + + (x ) + 8(y ) = 15 + x 6 + 8y 4 = 15 + x + 8y, as necessary. Therefore P = (, ). 6. f x (x, y) = x + y, f y (x, y) = x y, and so L(x, y) = f(, ) + f x (, )(x ) + f y (, )(y ) = + ()() + [() + ()](x ) + [() ()](y ) = 1 + 1(x ). Hence z = f(.05,.96) f(, ) L(.05,.96) 1 = 1+1(.05 ) 1 = 1(0.05) = 0.65. 15

In general, z f x (x 0, y 0 ) x + f y (x 0, y 0 ) y. 7. The volume V of a cone with base radius r and height h is V (r, h) = πr h/. So V r (r, h) = πrh/, V h (r, h) = πr /, and by using the local linear approximation of V at the point (10, 5) gives relative error = true volume measured volume = V V r (10, 5) r + V h (10, 5) h V r (10, 5) r + V h (10, 5) h (10)(5)π (0.01) + (10) π (0.01) = 0π. Thus, the maximum relative error in the calculated volume is about 0π cm 6 cm. To get the maximum percentage error, recall that percentage error = true volume measured volume true volume = V true volume. We just established that V 0π cm, and since r, h 0.1 cm and the measured radius and height is 10 cm and 5 cm, respectively, we find that 9.9 cm true radius 10.1 cm and 4.9 cm true height 5.1 cm 16

so that π(9.9) (4.9) cm true volume π(10.1) (15.1) cm. Because we divide by the true volume to find percentage error, the possible percentage error is largest when the possible volume is at its smallest, and so percentage error = V true volume 0π π(9.9) (4.9) 0.046. Hence, the maximum percentage error is about.46%. 17