# Math 22 Calculus III Online - Summer 2013

## Homework Hints

Stewart WebAssign Description
13.1.3 13.1.1 For definition of xz-plane, etc, see paragraph preceding Example 2 on page 803
13.1.4 13.1.2 For definition of projection see top of page 802
13.1.9 13.1.3 Note, if the three points do NOT lie on a straight line then they form a triangle. The length of any one side of a triangle is less then the sum of the lengths of the other two sides
13.1.18 13.1.6 See Example 5. You can divide through by the coefficient of x2, y2 and z2
13.1.20 13.1.7 The center of the sphere would be the midpoint of the segment
13.1.22 13.1.8 Use the idea of Prob #4 to find the radius
13.1.39 13.1.10 Is X(x, y, z) is a point equidistant from A and B then |AX| = |BX|
13.2.24 13.2.6 First find a unit vector in the same direction as the given vector
13.3.28 13.3.8 If u = <x, y> is a solution then | u | = 1 AND u dot v = | u | | v | cos 600 = | v | cos 600. Solve these two equations simultaneously for x and y
13.4.28 13.4.8 Draw a picture to figure out which vectors to use
13.5.2 13.5.2 One needs a vector parallel to the line and a point on the line
13.5.4 13.5.3 One needs a vector parallel to the line and a point on the line
13.5.5 13.5.4 One needs a vector parallel to the line and a point on the line
13.5.7 13.5.5 One needs a vector parallel to the line and a point on the line
13.5.10 13.5.6 One needs a vector parallel to the line and a point on the line
13.5.12 13.5.7 The line of intersection lies in both planes so is orthogonal to the normal vectors of both planes
13.5.16 13.5.8 One needs a vector parallel to the line and a point on the line
13.5.24 13.5.10 One needs a vector normal (orthogonal) to the plane and a point on the plane
13.5.29 13.5.11 One needs a vector normal (orthogonal) to the plane and a point on the plane
13.5.31 13.5.12 One needs a vector normal (orthogonal) to the plane and a point on the plane
13.5.35 13.5.13 One needs a vector normal (orthogonal) to the plane and a point on the plane
13.5.55 13.5.17 Find vector parallel to the line of intersection
13.5.67 13.5.19 Draw a picture with vector from a point on the line to the given point and a vector from the same point on the line that is parallel to the line. Use dot or cross product to find length of appropriate side of an appropriate right triangle
13.5.69 13.5.20 Draw a picture with a vector from a point on the plane (that you find the coordinates for) to the given point and draw a vector normal to the plane to the given point. Use dot or cross product to find length of appropriate side of an appropriate right triangle
13.5.72 13.5.21 Find a point on one of the planes and then find distance from this point to the other plane as you did in the previous problem
13.6.ALL 13.6.All See the Exploring Multivariable Calculus Java Applet information
14.1.2 14.1.1 Recall the domain of logrithm functions is the set of positive real numbers
14.1.4 14.1.2 Recall L'Hopital's Rule
14.1.5 14.1.3 L'Hopital's Rule Twice
14.1.37 14.1.5 Solve for y and z in terms of x. You can check your solution using the Exploring Multivariable Calculus Java Applet
14.1.38 14.1.6 Solve for y and z in terms of x. You can check your solution using the Exploring Multivariable Calculus Java Applet
14.2.9 14.2.2 Product Rule
14.2.26 14.2.8 Need a point on the tangent line and a vector parallel to the tangent line.
14.2.32 14.2.9 Angles can be found using dot product
14.2.34 14.2.10 inverse tangent in j component, u-sub in k component
14.2.36 14.2.11 u-sub in j component, parts in k component
14.3.3 14.3.1 The quatity under the radical is a perfect square. First eliminate the negative exponent by writing as a fraction. Next get a common denominator finally recognize the numerator as a perfect square
14.3.6 14.3.2 The quatity under the radical is a perfect square
14.3.24 14.3.6 Find value of t so that r(t) equals the given point, then use Thm 10 with this t value instead of finding curvature for all t values
14.3.49 14.3.10 Two planes are parallel if the vectors perpendicular to them are parallel
14.4.19 14.4.5 To find the minimum (or maximum) of a function set the derivative of the function (here you need the function for the speed) equal to zero and find critical points
14.4.35 14.4.9 Tangential component of acceleration equals the component of a(t) in the direction of v(t). Use Pythagorean Thm to find normal component of acceleration
14.4.37 14.4.10 Tangential component of acceleration equals the component of a(t) in the direction of v(t). Use Pythagorean Thm to find normal component of acceleration
15.1.All 15.1.All Though calculators and computer graphics can be used to help answer these questions you should try to do all of them by hand just using your basic knowledge of properties of functions of a single variable. Only use technology to check that you have done the problem correctly
15.2.10 15.2.2 Recall, if y is close to 0 then sin(y) is close to y
15.2.11 15.2.3 If y is close to 0 then cos(y) is close to 1
15.2.14 15.2.4 Divide
15.2.16 15.2.5 Recall, if y is close to 0 then sin(y) is close to y
15.2.17 15.2.6 Rationalize the denominator
15.2.38 15.2.9 Note that this function is defined at every point in the x-y plane. This by itself does not say that the function is continuous at every point in the x-y plane.u^v = e^(v ln u)
15.3.22 15.3.4 u^v = e^(v ln u)
15.3.34 15.3.5 u^v = e^(v ln u)
15.4.4 15.4.1 See equation 2
15.4.6 15.4.2 See equation 2
15.4.17 15.4.3 f(x,y) = (2x+3)/(4y+1); See equation 2
15.4.40 15.4.10 delta(x) = delta(y) = delta(z) = delta(w) = 0.05
15.6.11 15.6.3 Remember the direction vector must be a unit vector
15.6.19 15.6.5 Find unit vector in the direction from P to Q
15.6.26 15.6.6 See Theorem 15
15.6.35 15.6.9 Draw a picture of the vectors
15.6.43 15.6.10 Subtract so the surface is a zero level surface of a function of three variables. Then the gradient of this function is perpendicular to the level surface
15.6.52 15.6.11 See hint for previous problem
15.7 Remember the Exploring Multivariable Calculus Java Applet gives an easy why to graph surfaces
15.7.8 15.7.1 Remember e^(whatever) is always greater than zero
15.7.12 15.7.2 If you graph this surface it looks like corregated tin
15.7.22 15.7.3 Don't forget the CHAIN RULE
15.7.23 15.7.4 Subtract df/dy from df/dx to get cos(x) = cos(y) which tells you that x = y + 2n*pi. But x is between 0 and 2*pi so x = y. Plug x in for y in df/dx = 0, and use double angle formula to get a quadratic equation in cos(x) which you can factor
15.7.30 15.7.5 If the derivative of f(x,y) restricted to a boundary curve is zero at all points in the curve, then the surface is level along that curve
15.7.34 15.7.7 Substitute for y^2 along the quarter circle part of the boundary
15.8.3 15.8.1 Solve x-component equation for x and substitute into y-component equation. Factor the resulting equation to get values for y and lambda. Back substitute into the component equations and the constraint equation to find x and y. Note: some will not work
15.8.5 15.8.2 Get x component equation equal to zero and factor. Substitute results into y-component equation of constraint equation to solve for x and y
15.8.8 15.8.3 The y-component equation gives values for lambda and y, but the x-component equation shows that lambda cannot equal zero. So you can solve the x and z-component equations for x and z in terms of lambda, respectively and substitute into constraint equation to solve for x and z
15.8.28 15.8.4 To simplify, note that the minimum distance will be at the same point as minimum distance squared so one can get rid of the square root. Solve the componenet equations for x, y, and z in terms of lambda and substitute into the constraint equation
15.8.30 15.8.5 Simplify by minimizing the SQUARE of the distance. Get y-component equation equal to zero and factor. Substitute into the remaining two component equations and the constraint equation
16.1.11 & 12 16.1.3 & 4 The value of the double integral will be to volume of the solid under the surface, see Fig 7
16.2.20 16.2.5 Integrate first with respect to y. Use a u-substitution, substituting u for the denominator
16.3.All 16.3.All Be sure to sketch the region over which you are integrating. Some of the early problems might could be completed without doing this, but it is the simple problems you practice and gain understanding so that you can do the harder problems later
16.4.All 16.4.All Be sure to sketch the region over which you are integrating
16.6.All 16.6.All Careful sketch becomes even more important for triple integrals. In general you will need a sketch of the solid region you are integrating over, and then a second sketch of the 2-dimensional region that you integrate over after completing the "inside" integral. The Exploring Multivariable Calculus Java Applet can help check that your 3-dimensional sketch is correct
16.7.All 16.7.All Carefully sketch. Use Exploring Multivariable Calculus Java Applet to check your sketch
16.8.All 16.8.All Carefully sketch. Use Exploring Multivariable Calculus Java Applet to check your sketch