| 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 |