260

If f(x, y) = 3 x3 + y3 , find fx(0, 9)

0

Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.) a = 5i − 8j + k, b = 6i − k

cos-1(29/3√370) 60 degrees

What does the equation x = 8 represent in r^2 and r^3

line, plane

Find the sum of the given vectors. a = 1, 0, 1 , b = 0, 5, 0

(1,5,1)

Find |u × v| and determine whether u × v is directed into the screen or out of the screen.

we have |u × v| = |u||v| sin(θ) = (7)(8) sin(45°) = 56 · √2 2 = 28 2 . By the right-hand rule, u × v is directed out of the screen.

Find the cross product a ⨯ b. a = 2, 5, 0 , b = 1, 0, 9

= (45 − 0)i − (18 − 0)j + (0 − 5)k = 45i − 18j − 5k Now (a ⨯ b) · a = 45, −18, −5 · 2, 5, 0 = 90 − 90 + 0 = 0 and (a ⨯ b) · b = 45, −18, −5 · 1, 0, 9 = 45 + 0 − 45 = 0, so a ⨯ b is orthogonal to both a and b.

Is the line through (−4, −6, 1) and (−2, 0, −3) parallel to the line through (4, 10, 13) and (3, 7, 15)?

Direction vectors of the lines are v1 = −2 − (−4), 0 − (−6), −3 − (1) = 2, 6, −4 and v2 = 3 − (4), 7 − (10), 15 − (13) = −1, −3, 2 , and since v2 = − 1 2 v1, the direction vectors and thus the lines are parallel.

Find the point at which the line intersects the given plane. x = 4 − 4t, y = 5t, z = 1 + t; x + 2y − z = 8

Substitute the parametric equations of the line into the equation of the plane: x + 2y − z = 8 right double arrow implies (4 − 4t) + 2(5t) − (1 + t) = 8 right double arrow implies 5t + 3 = 8 right double arrow implies t = 1. Therefore, the point of intersection of the line and the plane is given by x = 4 − 4(1) = 0, y = 5(1) = 5, and z = 1 + 1 = 2, that is, the point (0, 5, 2).

Find an equation of the tangent plane to the given surface at the specified point. z = ln(x − 9y), (10, 1, 0)

z = f(x, y) = ln(x − 9y) right double arrow implies fx(x, y) = 1/(x − 9y), fy(x, y) = −9/(x − 9y), so fx(10, 1) = 1, fy(10, 1) = −9, and an equation of the tangent plane is z − 0 = fx(10, 1)(x − 10) + fy(10, 1)(y − 1) right double arrow implies z = 1(x − 10) + (−9)(y − 1) or z = x − 9y − 1.

Let P be a point not on the line L that passes through the points Q and R. The distance d from the point P to the line L is d = |a × b| |a| where a = QR and b = QP. Use the above formula to find the distance from the point to the given line. (5, 2, −2); x = 1 + t, y = 3 − 2t, z = 5 − 3t

√195,14

Find a unit vector that is orthogonal to both i + j and i + k.

1/√3i-1/√3j-1/√3k

Find the work done by a force F = 4i − 8j + 9k that moves an object from the point (0, 8, 4) to the point (2, 12, 24) along a straight line. The distance is measured in meters and the force in newtons.

156 j

Find the distance from the point to the given plane. (1, −9, 6), 3x + 2y + 6z = 5

16/7

If a child pulls a sled through the snow on a level path with a force of 30 N exerted at an angle of 44° above the horizontal, find the horizontal and vertical components of the force. (Round your answers to one decimal place.)

21.6 20.8

Consider the points below. P(1, 0, 1), Q(−2, 1, 3), R(3, 2, 5) (a) Find a nonzero vector orthogonal to the plane through the points P, Q, and R. Find the area of the triangle PQR.

Because the plane through P, Q, and R contains the vectors PQ and PR, a vector orthogonal to both of these vectors (such as their cross product) is also orthogonal to the plane. Here PQ = −3, 1, 2 and PR = 2, 2, 4 , so PQ × PR = (1)(4) − (2)(2), (2)(2) − (−3)(4), (−3)(2) − (2)(1) = 0, 16, −8 Therefore, 0, 16, −8 or any nonzero scalar multiple thereof, such as 0, −16, 8 is orthogonal to the plane through P, Q, and R. (b) Note that the area of the triangle determined by P, Q, and R is equal to half of the area of the parallelogram determined by the three points. From part (a), the area of the parallelogram is PQ × PR = 0, 16, −8 = 0 + 256 + 64 = 320 = 8 5 , so the area of the triangle is 1 2 · 8 5 = 4 5 .

Determine whether the planes are parallel, perpendicular, or neither. 3x + 12y − 9z = 1, −12x + 24y + 28z = 0

Normal vectors for the planes are n1 = 3, 12, −9 and n2 = −12, 24, 28 , so the normals (and thus the planes) aren't parallel. But n1 · n2 = −36 + 288 − 252 = 0, so the normals (and thus the planes) are perpendicular.

A manufacturer has modeled its yearly production function P (the monetary value of its entire production in millions of dollars) as a Cobb-Douglas function P(L, K) = 1.47L0.65K0.35 where L is the number of labor hours (in thousands) and K is the invested capital (in millions of dollars). Find P(130, 55) and interpret it. (Round your answers to one decimal place.) P(130, 55) = (No Response) seenKey 141.4 , so when the manufacturer invests $(No Response) seenKey 55 million in capital and (No Response) seenKey 130 thousand hours of labor are completed yearly, the monetary value of the production is about $(No Response) seenKey 141.4 million.

P(130, 55) = 1.47(130)0.65(55)0.35 ≈ 141.4, so when the manufacturer invests $55 million in capital and 130 thousand hours of labor are completed yearly, the monetary value of the production is about $141.4 million.

Find parametric equations for the line. (Use the parameter t.) The line through the points 0, 1 2 , 1 and (8, 1, −8)

The vector v = 8 − 0, 1 − 1 2 , −8 −1 = 8, 1 2 , −9 is parallel to the line. Letting P0 = (8, 1, −8), parametric equations are x = 8 + 8t, y = 1 + 1 2 t, z = −8 − 9t, while symmetric equations are x − 8 8 = y − 1 1/2 = z + 8 −9 or x − 8 8 = 2y − 2 = z + 8 −9 .

Describe the surface in double-struck R3 represented by the equation x + y = 5

This is the set {(x, 5 − x, z) which is a vertical plane that intersects the xy-plane in the line y = 5 − x, z = 0.

Find a vector a with representation given by the directed line segment AB. A(0, 4, 5), B(2, 4, −4)

(2,0,−9)

Find the sum of the given vectors. a = −3, 6 b = 8, −2

(5,4)

Suppose you start at the origin, move along the x-axis a distance of 6 units in the positive direction, and then move downward along the z-axis a distance of 5 units. What are the coordinates of your position?

(6,0,-5)

Let g(x, y) = cos(x + 5y). (a) Evaluate g(10, −2). g(10, −2) Find the domain of g. Find the range of g. (Enter your answer using interval notation.

(a) g(10, −2) = cos(10 + 5(−2)) = cos(0) = 1 (b) x + 5y is defined for all choices of values for x and y and the cosine function is defined for all input values, so the domain of g is the set of real numbers2. (c) The range of the cosine function is [−1, 1] and x + 5y generates all possible input values for the cosine function, so the range of cos(x + 5y) is [−1, 1].

Write the equation of the sphere in standard form. x2 + y2 + z2 + 2x − 2y − 2z = 22 Find its center and radius

(x+1)^2+(y-1)^2+(z-1)^2=25 -1,1,1 5

Find an equation of the sphere with center (−5, 3, 6) and radius 6. What is the intersection of this sphere with the yz-plane?

(x+5)^2 +(y-3)^2 +(z-6)^2 (y-3)^2+(z-6\^2=11

Find the first partial derivatives of the function. f(x, y) = x8 + 6xy3

(x, y) = x8 + 6xy3 right double arrow implies fx(x, y) = 8x7 + 6y3, fy(x, y) = 0 + 6x · 3y2 = 18xy2

Find an equation of the sphere that passes through the origin and whose center is (2, 1, 2).

(x-2)^2+(y-1)^2(z-2)^2=9

If f(x, y) = 9 − 3x2 − y2 , find fx(9, 6) and fy(9, 6) and interpret these numbers as slopes.

-54 -12

Find a · b. a = 2i + j, b = i − 8j + k

-6

ind the vector, not with determinants, but by using properties of cross products. (i × j) × k

0

Determine whether the given vectors are orthogonal, parallel, or neither. (a) a = 9, 3 , b = −2, 6 orthogonal parallel neither Correct: Your answer is correct. (b) a = 8, 5, −2 , b = 3, −1, 5 orthogonal parallel neither Correct: Your answer is correct. (c) a = −12i + 4j + 8k, b = 9i − 3j − 6k orthogonal parallel neither Correct: Your answer is correct. (d) a = 2i − j + 2k, b = 5i + 6j − 2k orthogonal parallel neither Correct: Your answer is correct.

1 3 2 1

Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (4, −7, 2) and parallel to the vector 1, 3, − 2 3

2 For this line, we have r0 = 4i−7j+2k and v = i+3j− k, so a vector equationisr=r0+tv=(4i−7j+2k)+ti+3j− k 2 3 2 3 = (4+t)i+(−7+3t)j+ 2− 3t k and parametric equations are x = 4+t, 2 y = −7 + 3t, z = 2 − 3 t.

Find a · b. |a| = 7, |b| = 6, the angle between a and b is 30°.

21√3

Find a · b. a = 4, 1, 1/5 b = 8, −5, −10

25

consider the following planes. x + y + z = 3, x + 8y + 8z = 3 (a) Find parametric equations for the line of intersection of the planes. (Use the parameter t.) (b) Find the angle between the planes. (Round your answer to one decimal place.)

3,-7t,7t 30.2

ind an equation of the plane. The plane through the point (5, 3, 2) and with normal vector 3i + j − k

3x+y-z=16

Find the lengths of the sides of the triangle PQR. P(5, −2, −3), Q(9, 0, 1), R(11, −4, −3)

6 6 2√10

Find the scalar and vector projections of b onto a. a = −5, 12 b = 6, 9

6, -30/13 72/13

The length and width of a rectangle are measured as 36 cm and 30 cm, respectively, with an error in measurement of at most 0.1 cm in each. Use differentials to estimate the maximum error in the calculated area of the rectangle.

6.6

Find a unit vector that has the same direction as the given vector. 8i − j + 4

8/9i -1/9j+4/9k

Find the volume of the parallelepiped with adjacent edges PQ, PR, and PS. P(−2, 1, 0), Q(4, 2, 4), R(1, 4, −1), S(3, 6, 2)

a = PQ = 6, 1, 4 , b = PR = 3, 3, −1 , and c = PS = 5, 5, 2 . a · (b × c) = 6 1 4 3 3 −1 5 5 2 = 6 3 −1 5 2 − 1 3 −1 5 2 + 4 3 3 5 5 = 66 − 11 + 0 = 55, so the volume of the parallelepiped is 55 cubic units.

Find the volume of the parallelepiped determined by the vectors a, b, and c. a = 1, 3, 3 , b = −1, 1, 2 , c = 4, 1, 2

a · (b × c) = 1 3 3 −1 1 2 4 1 2 = 1 1 2 1 2 − 3 −1 2 4 2 + 3 −1 1 4 1 = 1 · (2 − 2) − 3 · (−2 − 8) + 3 · (−1 − 4) = 15.

If a = 2, −1, 7 and b = 2, 2, 1 , find the following. a × b b*a

a × b = i j k 2 −1 7 2 2 1 = −1 7 2 1 i − 2 7 2 1 j + 2 −1 2 2 k = (−1 − 14)i − (2 − 14)j + [4 − (−2)]k = −15i + 12j + 6k b × a = i j k 2 2 1 2 −1 7 = 2 1 −1 7 i − 2 1 2 7 j + 2 2 2 −1 k = [14 − (−1)]i − (14 − 2)j + (−2 − 4)k = 15i − 12j − 6k

Find a + b, 9a + 3b, |a|, and |a − b|. a = 9i − 4j + 3k, b = 6i − 9k

a+b=15i-4j-6k 9a+3b=99i-36j a=√106 a-b=13

A(−4, 0, −9), B(2, 2, −3), C(1, 1, 2) Which of the points is closest to the yz-plane? Which point lies in the xz-plane?

c a

Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.) a = 2, 3 b = 5, −1

cos⁻1(7/√13/2) 68 degrees

Find the first partial derivatives of the function. f(x, y) = x y

f(x, y) = x y = xy−1 right double arrow implies fx(x, y) = y−1 = 1 y , fy(x, y) = −xy−2 = − x y2

Find all the second partial derivatives. f(x, y) = x4y − 4x3y2

f(x, y) = x4y − 4x3y2 right double arrow implies fx(x, y) = 4x3y − 12x2y2, fy(x, y) = x4 − 8x3y. Then fxx(x, y) = 12x2y − 24xy2, fxy(x, y) = 4x3 − 24x2y, fyx(x, y) = 4x3 − 24x2y, and fyy(x, y) = −8x3.

Find an equation of the sphere that passes through the point (5, 1, −5) and has center (3, 6, 3).

r= (3−5)2 +(6−1)2 +[3−(−5)]2 =√93. Thus,anequationofthesphereis(x−3)2 +(y−6)2 +(z−3)2 =93.

Find an equation of the plane. The plane through the points (0, 1, 1), (1, 0, 1), and (1, 1, 0)

x+y+z=2

Use implicit differentiation to find ∂z/∂x and ∂z/∂y. x2 + 8y2 + 9z2 = 3

x2 + 8y2 + 9z2 = 3 right double arrow implies ∂ ∂x (x2 + 8y2 + 9z2) = ∂ ∂x (3) right double arrow implies 2x + 0 + 18z ∂z ∂x = 0 right double arrow implies 18z ∂z ∂x = −2x right double arrow implies ∂z ∂x = −2x 18z = − x 9z , and ∂ ∂y (x2 + 8y2 + 9z2) = ∂ ∂y (3) right double arrow implies 0 + 16y + 18z ∂z ∂y = 0 right double arrow implies 18z ∂z ∂y = −16y right double arrow implies ∂z ∂y = −16y 18z = − 8y 9z .

Find the scalar and vector projections of b onto a. a = 4, 7, −4 b = 4, −1, 1

|a| = 42 + 72 + (−4)2 = 81 = 9 so the scalar projection of b onto a is compab = a · b |a| = (4)(4) + (7)(−1) + (−4)(1) 9 = 5/9 . The vector projection of b onto a is projab = a · b |a| a |a| = 5 9 · 1 9 4, 7, −4 = 5 81 4, 7, −4 = 20/81 , 35/81 , − 20/81 .