#$&*
course Mth 277
4/6 8Resubmit because forgot to add access code before
11.6
*********************************************
Question: `q001. Find grad(f) when f(x,y,z) = e^(x+y+z).
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
grad(f) for f(x,y,z) = e^(x+y+z)
f_x(x,y,z)= d/dx(e^(x+y+z)) = 1* e^(x+y+z) = e^(x+y+z)
f_y(x,y,z)= d/dy(e^(x+y+z)) = 1* e^(x+y+z) = e^(x+y+z)
f_z(x,y,z)= d/dz(e^(x+y+z)) = 1* e^(x+y+z) = e^(x+y+z)
grad(f) = (e^(x+y+z))i + (e^(x+y+z))j + (e^(x+y+z))k
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: grad(f) = del f = e^(x + y + z) i + e^(x + y + z) j + e^(x + y + z) k.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
Is the del f just another way of writing grad(f)?
@& Right. The 'del' operator is just the upside-down Delta, which is also written boldfaced to indicate that it's a vector operator.
del f = f_x `i + f_y `j + f_z `k.*@
------------------------------------------------
Self-critique rating:
*********************************************
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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
grad f(x,y) = x^2+ xy
f_x(x,y)= d/dx(x^2+ xy) = 2x + y f_x(1,-1) = 2*(1) + (-1) = 1
f_y(x,y)= d/dy(x^2+ xy) = x f_y(1,-1) = (1) = 1
grad f(1,-1) = (1)i + (1)j = i+j
????A unit vector in the direction of v is????
u = (v/||v||) = (i-j)/(`sqrt(1^2 + (-1)^2) = 1/(`sqrt(2))*(i-j) =`sqrt(2)/2*(i-j)
Thus,
D_u(x,y) = grad f(x,y)*u = (1)*(1/(`sqrt(2)) - (1) *(1/(`sqrt(2))
= 0
????I’m going to leave it as is so you see my mistake, but I know I am to get the dot product and not mult the gradient and unit vector?????
Confidence rating:
.............................................
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).
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
So really the steps are
1)Find grad f(x,y)
2)Find unit vector of v vector
3)then find dot product of grad and unit vector
The final result is the directional derivative in the direction of v.
????I see the steps and know what I’m doing with the formulas but still not quite positive about everything that is going on???????
@& In general, suppose the unit vector is
`u = a `i + b `j = cos(theta) `i + sin(theta) `j.
f_x tells you how fast the value of f changes if you move in the x direction.
a (or cos(theta)) is how far you move in the x direction if you move 1 unit in the direction of `u.
So if you move the direction of `u, you move a units in the x direction for every unit in the `u direction, and this x motion causes the value of f to change at the rate a * f_x.
Similarly motion in the direction of `u results in some motion in the y direction, moving b units in the y direction for every unit moved in the direction of `u. This y motion causes an additional change in the value of f, which changes at the rate b * f_y.
The change in f, per unit of distance moved in the direction of `u, is therefore f_x * a + f_y * b.
The rate of change in f, per unit of distance in the direction of `u, is the directional derivative of f in the direction of `u.
This is equal to the dot product of the gradient and the vector `u:
grad f dot `u = (f_x `i + f_y `j) dot (cos(theta) `i + sin(theta) `j).
For example if u is it angle 37 degrees with the x direction, we have
`u = .8 `i + .6 `j
so that
grad f dot `u = .8 * f_x + .6 * f_y.
This means that when you move in the direction of `u, you move 80% as far in the x direction and 60% as far in the y direction, so the rate at which the value of f changes with respect to distance is 80% of f_x (the rate of change in the x direction) plus 60% of f_y (the rate of change in the y direction).
*@
------------------------------------------------
Self-critique rating:
*********************************************
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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Re-writing so that gradient theorem applies, we let F(x,y,z) = x^3 + 2xy^2 + 3y - z and let S be the level surface F(x,y,z) = 2. The gradient grad F is normal to S at P_0 = (1,1,1)
grad F(x,y,z) = (3x^2+2y^2)i + (4xy+ 3)j + (-1)k
Normal vector at P_0 is
N = grad F_0 = grad F(1,1,1) = 5i + 7j - k
Tang plane is
5(x-1) + 7(y-1) - (z-1) = 0 or 5x + 7y - z = 11
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
Again really good at using formulas but not seeing really clearly what’s going on and the reason everything works out like it does.
@& S is a level surface of f(x, y, z). This means that as you move along S the value of the function doesn't change. So if you move a little ways from a point of the surface, in any direction along the tangent plane at that point, you will at first encounter 0 rate of change.
The gradient, on the other hand, is the direction in which the value changes most quickly.
A direction in which there is no change is perpendicular to the direction in which the change is most rapid.
Thus any vector perpendicular to the gradient will be in a direction of no change.
It follows that the gradient is normal to the tangent plane, so the tangent plane at (x0, y0, z0) is the plane through the point (x0, y0, z0) which is normal to f_x `i + f_y `j + f_z `k. That plane is characterized by
( (x - x0) `i + (y - y_0) `j + (z - z0) `k) dot (f_x `i + f_y `j + f_z `k) = 0.
*@
------------------------------------------------
Self-critique rating:
*********************************************
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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
grad(f) for f(x,y,z) = z ln (y/x)
f_x(x,y,z)= d/dx(z ln (y/x)) = z*(d/dx ln(y/x))
f(g(h))’ f(z)=ln(z), g(h)=(y/x), f’(z)=1/(z), g(h)= -y/x^2
d/dx(z ln (y/x))=z*((1/(y/x))*(-y/x^2) = z(-yx/yx^2)
= -z/x
f_y(x,y,z)= d/dy(z ln (y/x)) = z*(d/dx ln(y/x))
f(g(h))’ f(z)=ln(z), g(h)=(y/x), f’(z)=1/(z), g(h)= -y/x^2
d/dx(z ln (y/x))=z*((1/(y/x))*(1/x) = z(x/yx)
= z/y
f_z(x,y,z)= d/dz(z ln (y/x)) = ln(y/x)
grad(f) = (-z/x)i + (z/y)j + (ln(y/x))k
At (1,e,-1), grad(f) =(1)i + (-1/e)j + (ln(e))k
= i +(-1/e)j + k
Most rapid increase is || grad(f)||
|| grad(f)|| = `sqrt(1^2 + (-1/e)^2+1^2) = `sqrt(1+(1/e^2)+1)
=` sqrt( (1/e^2) + 2)
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
Messed up algebra somewhere but you answer was correct where -1/e was not for j component. I also didn’t compute the unit vector but understand it is simply the gradient divided by the greatest rate of increase. The greatest rate of decrease is -` sqrt(2 + 1/e^2) right????
@& The rate of change in the direction of the gradient is the magnitude of the gradient, and that is the greatest rate of change possible at that point.
The greatest rate of decrease is the negative of the greatest rate of increase.
In other words, you're right.*@
------------------------------------------------
Self-critique rating:
*********************************************
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).
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Having a hard time getting started going to look at the solution.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I think I understand what just happened, I mean how you arrived at your solution. Just one question is r = (x)i + (y)j + (z)k, so that r^2 = (x^2)i + (y^2)j + (z^2)k ????
@& Let `r be the vector and r its magnitude.
`r = x `i + y `hj + z `k
so
r = || r || = sqrt( x^2 + y^2 + z^2)
It follows that
r^2 = x^2 + y^2 + z^2.
r^2 is not r^2 = (x^2)i + (y^2)j + (z^2)k, among other things because r^2 is not a vector.*@
------------------------------------------------
Self-critique rating:
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
&#This looks good. See my notes. Let me know if you have any questions. &#