LB 220 Homework 4 Solutions Section 11.4, # 40: This problem was solved in class on Feb. 03. Section 11.4, # 42: This problem was also solved in class on Feb. 03. Section 11.4, # 43: Also solved in class on Feb. 03. Section 11.5, # 2: Solved in class on Feb. 03. Section 11.5, # 10: Find the directional derivative of f(x, y) = 2x 2 + y 2 at ( 1, 1) in the direction of A = 3 i 4 j. Compute the gradient: f = 4x i + 2y j, evaluate at ( 1, 1): f( 1, 1) = 4 i + 2 j, and scale A to the unit vector u = 3/5 i 4/5 j (since A = 5). The directional derivative is f( 1, 1) u = 12/5 8/5 = 20/5 = 4. Section 11.5, # 26: Sketch f(x, y) = 7, where f(x, y) = x 2 xy + y 2, together with f and the tangent line at ( 1, 2). Also, write an equation for this tangent line. Since f = (2x y) i + ( x + 2y) j, f( 1, 2) = 4 i + 5 j. This vector is normal to the level curve f = 7. To obtain a tangent vector, you can either use the fact that this vector is parallel to lines of slope ( 5/4), or you could directly try to find a tangent vector T by algebraically determining a solution to the equation T f( 1, 2). The first approach is easier since it is clear that by taking the negative reciprocal of slope that the tangent line has slope (4/5). But this method does not work if the vector was a vector in three dimensions. So, I ll give an example of the latter approach since it works in all dimensions: The vector T is unknown. Assign variables to the unknown components (two variables in this case, three variables in three dimensions, etc.): let T = a i + b j. We are to solve the equation (a i + b j) ( 4 i + 5 j) = 0
This reduces to 4a + 5b = 0. This is an underdermined system (of one equation and two variables), so we are free to choose one of the variables. I ll chose a = 1. This means that 4 + 5b = 0, and so b = 4/5. So a tangent vector is i + (4/5) j; and this vector has slope (4/5). (What would happen if I had chosen a = 2 instead? Will we still obtain the same slope?) Therefore, the tangent line has slope 4/5 and has equation (y 2) = (4/5)(x + 1). Here is the sketch of the level curve f = 7 together with the gradient vector at the point ( 1, 2): Section 11.5, # 32: Suppose that the derivative of f(x, y, z) at P is greatest in the direction of v = i + j k; and that, in this direction, the derivative is 2 3. What is f(p )? What is the derivative of f at P in the direction of i + j?
Let u = (1/ 3)v, the unit vector in the direction of v. We are given that D u f(p ) = 2 3. We also know that the direction in which the gradient is greatest is that of the gradient. And we know that the length of the gradient is the value of the derivative in this direction. This means that f(p ) has direction u and magnitude 2 3. Hence, f(p ) = 2 3 u = 2 i + 2 j 2 k. We are also asked to compute the directional derivative of f in the direction (1/ 2)(i + j). This has value f(p ) (1/ 2)(i + j) = (2 i + 2 j 2 k) (1/ 2)(i + j) = 2 2 Section 11.6, # 2: Find the equation of the tangent plane and the normal line to the surface x 2 + y 2 z 2 = 18 at the point P = (3, 5, 4). Let g(x, y, z) = x 2 + y 2 z 2. The surface is the level surface g = 18. Therefore, g(p ) is normal to the surface at P. If n is a non-zero vector and P = (a, b, c) is a point, then the equation of the plane containing P and having normal vector n has vector equation n P Q = 0, where Q = (x, y, z). In coordinates, this is A(x a) + B(y b) + C(z c) = 0, where n = A i + B j + C k. (This is expained in section 9.5, p. 532.) Therefore, for the problem at hand, we can use n = g(p ) and P = (3, 5, 4). The gradient of g is g = 2x i + 2y j 2z k Evaluated at P, this vector is 6 i + 10 j + 8 k. Thus, an equation for the tangent plane is 6(x 3) + 10(y 5) + 8(z + 4) = 0. The normal line will have vector equation f(t) = OP + tn, where O is the origin. Thus, the normal line is given by the vector valued function f(t) = (3 i + 5 j 4 k + t(6 i + 10 j + 8 k) =
(3 + 6t) i + (5 + 10t) j + ( 4 + 8t) k. This parametrization by a single vector valued function is equivalent to the parametrization by three scalar functions: x(t) = 3 + 6t, y(t) = 5 + 10t, z(t) = 4 + 8t. Section 11.6, # 10: Find an equation for the tangent plane to the surface z = exp( x 2 y 2 ) at the point P = (0, 0, 1). Let g(x, y, z) = exp( x 2 y 2 ) z. Then the surface is the level surface g = 0. Therefore, g(p ) is normal to the surface at P. The gradient of g is g = 2xe x2 y 2 i 2ye x2 y 2 j k. Thus, g(p ) = k. And so the tangent plane is 0(x 0) + 0(y 0) 1(z 1) = 0 which simplifies to z = 1. Section 11.6, # 14: Find parametric equations for the line tangent to the curve which is the intersection of the surfaces xyz = 1 and x 2 + 2y 2 + 3z 2 = 6 at the point P = (1, 1, 1). Strategy: If f = xyz and g = x 2 + 2y 2 + 3z 2, then both surfaces are level surfaces: f = 1 and g = 6, respectively. Therfore f(p ) and g(p ) are normal to the curve of intersection. And, hence, the cross product of these two vectors will be tangent to the curve. f = yz i + xz j + xy k, g = 2x i + 4y j + 6z k, f(p ) = i + j + k g(p ) = 2 i + 4 j + 6 k The cross product of f(p ) and g(p ) is the determinant of i j k 1 1 1, 2 4 6 which equals 2 i 4 j + 2 k. And so, the following are parametric equations for the line tangent to the curve of intersection of these two surfaces: x(t) = 1 + 2t, y(t) = 1 4t, z(t) = 1 + 2t.
Section 11.6, # 24: The temperature in degrees Celsius of a region in in space is given by T (x, y, z) = 2x 2 xyz. A particle moves in this region so that its position (in meters) is described as a function of time t measured in seconds by the parametric equations x = 2t 2, y = 3t, and z = t 2. How fast is the temperature of the particle changing in degrees Celsius per meter when the particle is at P = (8, 6, 4)? How fast is the temperature changing at P in degrees Celsius per second? The first question asks for a computation of the directional derivative of T in the direction of the velocity vector of the particle. The second question asks for a computation of the derivative of T with respect to t; this is the same thing as the dot procut of the gradient of T and the velocity vector of the particle at P. So, T (P ) = 56 i + 32 j 48 k. The velocity of the particle is T = (4x yz) i xz j xy k. v(t) = 4t i + 3 j 2t k. At P, t = 2, so the velocity at P is equal to v(2) = 8 i + 3 j 4 k. The unit vector in the direction of v(2) is u = (1/ 89) v(2). So, the rate of change of T in units of degrees Celsuius per meter is D u T = T (P ) u = (1/ 89)(448 + 96 + 192) = 734/ 89 77.8. And the rate of change of the temperature at P in degrees Celsius per second is T (P ) v(2) = 734. Section 11.6, # 36: f(x, y) = ln x + ln y. Determine the linearization L(x, y) of f at P = (1, 1) and determine an upper bound for the error in the approximation f(x, y) L(x, y) over the rectangle x 1 0.2, y 1 0.2.
Since f x = 1/x and f y = 1/y, the linearization at P is L(x, y) = f(1, 1)+f x (1, 1)(x 1)+f y (1, 1)(y 1) = 0+(x 1)+(y 1) = x+y 2. The error is less than or equal to (M/2)(0.2 + 0.2) 2 (see p. 630 of the textbook), where M is greater than or equal to the absolute value of the second partial derivatives f xx, f xy, and f yy restricted to the rectangle. f xx = 1/x 2, f xy = 0, and f yy = 1/y 2. In absolute value and restricted to the rectangle in question, the largest that any of these could be is 1/(0.8) 2 = 100/64. So, an upper bound for the error is (100/128)(0.4) 2 = 16/128 = 1/8 = 0.125. For example, f(1.1, 0.9) (1.1) + (0.9) 2 = 0 with an error of at most 0.125. This means that ln 1.1 + ln 0.9 is between 0.125 and 0.125.