Query116mth

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course Mth 277

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Question: `q001. Find grad(f) when f(x,y,z) = e^(x+y+z).

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Your solution:

f_x= 1e^(x+y+z)

f_y= 1e^(x+y+z)

f_z= 1e^(x+y+z)

grad(f)=[e^(x+y+z)]i+[e^(x+y+z)]j+[e^(x+y+z)]k

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Given Solution: grad(f) = del f = e^(x + y + z) i + e^(x + y + z) j + e^(x + y + z) k.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: `q002. Find the directional derivative of f(x,y) = x^2 + xy at the point (1, -1) in the direction of the vector v = i - j.

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Your solution:

f_x=2x+y

f_y=x

D_v f(1,-1)=f_x(1,-1)(1)+f_y(1,-1)(-1)

=[2(1)+(-1)](1) + [1](1)

=1-1=0

Therefore, the directional derivative is 0.

confidence rating #$&*:

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Given Solution:

grad(f) = del f = (2 x + y) i + x j.

At (1, -1) the gradient is therefore i + j.

The unit vector in the direction of v is sqrt(2) / 2 * (i - j).

The directional derivative in the direction of v is the dot product of the gradient and the unit vector.

In this case the directional derivative is zero. At the point (1, -1) the direction of the vector v is the one in which f(x, y) has zero rate of change. v would be tangent to a level curve at (1, -1).

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: `q003. Find a unit vector which is normal to the surface given by the equation 2 = x^3 + 2xy^2 + 3y - z at the point P = (1,1,1). Also find the equation of the tangent plane at this point using this information.

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Your solution:

2 = x^3 + 2xy^2 + 3y - z

P = (1,1,1)

grad(f)=f_x i + f_y j +f_z k

f_x=3x^2+2y^2 i

f_y=4yx+3 j

f_z=-1 k

Now I will plug in point P(1, 1, 1) at the deriv in respect of the each var.

3(1)+2(1)=5i

4(1)(1)+3=7j

-1=-k

Unit vect normal to the surface given by the equation is 5i+7j-1k

To find tang plane, we do the dot prod to = 0

5(x-1)+7(y-1)-1(z-1)=0

5x+7y-z=11

confidence rating #$&*:

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Given Solution: The gradient of f(x, y, z), evaluated at a point, is normal to the surface.at that point.

The gradient is easily found to be

grad f = (3 x^2 + 2 y^2) i + (4 x + 3) j - k. At (1, 1, 1) the gradient is therefore (5 i + 7 j - k).

The tangent plane passes through (1, 1, 1) and is normal to 5 i + 7 j - k.

A point (x, y, z) lies on the tangent plane if the vector (x - 1) i + (y - 1) j + (z - 1) k is perpendicular to the normal vector. So (x, y, z) lies on the plane if the dot product of this vector and the normal vector is zero.

The dot product is 5 (x - 1) + 7 ( y - 1) - (z - 1), so the equation of the tangent plane is

5 (x - 1) + 7 ( y - 1) - (z - 1) = 0, which simplifies to

5 x + 7 y - z - 11 = 0.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: `q004. Find the direction from the point P = (1,e,-1) in which the function f(x,y,z) = z ln (y/x) increases the most rapidly and compute the magnitude of the greatest rate of increase.

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Your solution:

P = (1,e,-1)

f(x,y,z) = z ln (y/x)

f_x=(-z/x)

f_y=(z/y)

f_z=ln (y/x)

[-z/x]i + [-z/y]j + [ln (y/x)]k

(1,e,-1):[1/1]i + [1/e]j + [ln (e/1)]k

i + (1/e)j + k

unit vect

i/(sqrt(2+(1/e)^2))+j(1/e)/(sqrt(2+(1/e)^2))+k/(sqrt(2+(1/e)^2))

.684i+.252j+.684k

magn=.9996

@&

This is the magnitude of the unit vector, which is in fact 1. The difference .0004 is a rounding error.

*@

confidence rating #$&*:

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Given Solution: The gradient is -z/x i + (z/y) j + ln(y/x) k.

At (1, e, -1) we get i + 1/e j + k.

A unit vector in this direction is u = i / sqrt(2 + 1/e^2) + j ./ (e sqrt(2 + 1/e^2)) + k / sqrt(2 + 1/e^2), approximately .68 i + .25 j + .68 k.

The magnitude of this vector is sqrt(2 + 1/e^2). The magnitude of the gradient is the greatest rate of increase.

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Self-critique (if necessary):

I think I did well untill I came to the magnitude of the vector. I probably should not of used the rounded values because my answer was considerably off. However, it seemed to far to be a rounding error so Im not sure exactly what happened.

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Self-critique rating:OK

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Question: `q005. A particle P1 with mass m1 is located at the origin, and a particle P2 with mass 1 unit is located at the point (x,y,z). According to Newton's law of universal gravitation, the force P1 exerts on P2 is modeled by F = -G(m1(xi + yj + zk))/r^3 where r is the distance between P1 and P2 and G is the gravitational constant.

Starting from the fact that r^2 = x^2 + y^2 + z^2, show that d/dx*(1/r) = -x/r^3, d/dy*(1/r) = -y/r^3 and d/dz(1/r) = -z/r^3. (Here d/dx denotes partial with respect to x)

The function V = -G*m1/r is called the potential energy function for the system. Show that F = -grad(V).

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Your solution:

r^2 = x^2 + y^2 + z^2

r=sqrt(x^2 + y^2 + z^2)

1/r=sqrt(1/x^2 + 1/y^2 + 1/z^2)

d/dx*(1/r) = -x/r^3, d/dy*(1/r) = -y/r^3

d/dz(1/r) = -z/r^3

F = -G(m1(xi + yj + zk))/r^3

V = -G*m1/r

This is where I got confused.

confidence rating #$&*:

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Given Solution: 1/r = 1 / sqrt(x^2 + y^2 + z^2) = (x^2 + y^2 = z^2)^(-1/2), so d/dx (1/r) = 2 x * (-1/2) (x^2 + y^2 + z^2)^(-3/2) = -x / (sqrt(x^2 + y^2 + z^2)) ^3 = -x / r^3.

The results for the y and z derivatives are acquired by a completely analogous series of steps.

It follows that the gradient of V = - G m1 / r is

V_x i + V_y j + V_z k = -G m1 * ( -x / r^3 i - y / r^3 j - z / r^3 k ). Simple rearrangement shows that this is identical to the force function F.

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Self-critique (if necessary):

I see where I got to and how I had to continue to relate things to achieve the final answer. It would take a little bit of rearanging, but it is obvious that these functions are identical.

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Self-critique rating:OK"

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&#Your work looks good. See my notes. Let me know if you have any questions. &#