Carol Said:

Calculus Instructor Solutions manual needed for Homeschool?

We Answered:

call the editors, they will send you the complete solution manual, or instructors manual for a fee .
try e-bay...
but the best thing is not to cheat... you do not need the answers to all the problems, if you understand and you are learning, with the solutions to the odd problems is MORE than enough... plus youhave the examples that are worked in the text book... .

Eileen Said:

Stuck on minimum and maximum problem multivariable calculus.?

We Answered:

Did you use the method of Lagrange multipliers? If so, the equations you should have gotten are

x(1 - 2?x²) = 0
y(1 - 2?y²) = 0
z(1 - 2?z²) = 0

This represents 8 possibilities, but we have to exclude x = y = z = 0 because that doesn't satisfy the constraint.

3 of the possibilities take two of the variables to be zero; then the third variable is ±1. All of these make f = 1, which I agree would be the minimum.

If we let exactly one of the variables be zero (3 possibilities), for example, x = 0, y ? 0, z ? 0, then we must have

1 - 2?y² = 0
1 - 2?z² = 0

giving

1/(2?) = y² = z². (so y = ± z)

Setting x = 0 and substituting for, say, y² in the constraint, we get

2z^4 = 1

So z = ± (½)^(¼). This gives the four solutions

(0, (½)^(¼), (½)^(¼))
(0, -(½)^(¼), (½)^(¼))
(0, (½)^(¼), -(½)^(¼))
(0, -(½)^(¼), -(½)^(¼))

all of which give f = ?2

You can let y or z be the one zero solution; you'll get solutions for the remaining variables much like the above, all of them giving f = ?2

Finally, there is the choice that none of them is zero. Then we get

x² = y² = z² = 1/(2?) Then

x^4 = y^4 = z^4.

By the constraint,

x^4 + x^4 + x^4 = 1
x = ±(1/3)^(1/4)

Again, a bunch of solutions, but they all lead to f = ?3

Francis Said:

Multivariable Calculus help? Please? :D?

We Answered:

will do 2 & 3....let p_w[h]= partial of h with respect to w
#2. int over the disk of [x^2 + z^2] sqrt{ 1 + (p_x[y])^2 + (p_z[y])^2} dA(x,z) = int [x^2 + z^2] sqrt { 1 + 4x^2 + 4z^2} dA...to polar coords....int over r in [0,2] & theta in [0,2 pi] of r^2 sqrt{1 + 4r^2} r dr dtheta ...let w = 1 + 4 r^2, w in [1,5]..you can do the calculations.
#3. "del" = < p_x [ ], p_y[ ], p_z [ ]>
curl F= "del' x F = < x-y , 0 - y , 1 - 0>...div F = "del" dot F = 0 + z - [sqrt z] / 2

Ronald Said:

in multivariable calculus,what number c are the vector<c,2> and <c,8> perpendicular?

We Answered:

The inner product (also called dot product) must be zero, i.e. c^2 + 16 = 0. This has no solution in real numbers so the two vectors are not perpendicular for any value of c.

Janice Said:

find the volume? In multivariable calculus?application of multiple integrals?

We Answered:

2) z1=x²+3y² and z2=4-y²
z1 and z2 intersect at z1=z2 ==>
x²+3y²=4-y²
x²+4y²=4
x²/4+y²=1 (ellipse)

A: x²/4+y²?1, dA=dxdy
Substitution with polar coordinates:
x=2rcos?, y=rsin?
(2rcos?)²/4+(rsin?)²?1 ==> r²?1, r?1
A1: 0???2? and 0?r?1
J=2r, dA=J*dA1, dxdy=J*drd?=2r*drd?


V=??[A] (z2-z1) dA=
??[A] (4-x²-4y²) dA=
??[A1] [4-(2rcos?)²-4(rsin?)²] J*dA1=
??[A1] [4-4r²]*2r*dA1=

2?. 1
8?d? ?(r-r³)dr=16?(r²/2-r^4/4) [r=0 to 1]=
.0 . 0

16?(1/2-1/4)=4?
......................................…
1) z1=0, z2=y
y=x² intersects with y=4 at x=±2
A: -2?x?2 and x²?y?4

V=??[A] (z2-z1) dA=??[A] y dxdy=

2 ? 4
? ? ?y dydx=
-2 x²
. ?
2? (y²/2) dx [y=x² to 4] =
.?
?
? (16-x^4) dx =(16x-x^5/5) [x=0 to 2] =
?
32-32/5=32(1-1/5)=128/5=25.6

Calvin Said:

find the volume?In multivariable calculus?application of multiple integrals?

We Answered:

Much Interesting question!!!!