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course Mth 277
November 15 around 7pm.
Query 11.6*********************************************
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):
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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
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^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):
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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:
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
Plug in point P(1, 1, 1)
3(1)+2(1)=5
4(1)(1)+3=7
Unit vector normal to surface=5i+7j-1k
Tangent plane:
5(x-1) + 7(y-1) – 1(z-1)
or
5x+7y-z=11
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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):
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Question: `q004. Find the direction from the point P = (1,e,-1) 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:
Du of f(x,y,z) at P(1,e,-1) in direction of the unit vector “u” is given by:
D_u f= grad(f0)*u
f_x=(-z/x)
f_y=(z/y)
f_z=ln (y/x)
[-z/x]i + [z/y]j + [ln (y/x)]k then plug in points (1,e,-1)
[1/1]i + [-1/e]j + [ln (e/1)]k then simplify
i + (-1/e)j + k
Direction of this:
Sqrt of (1 + (-1/e)^2)
Where do I go from here?
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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):
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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 tha F = -grad(V).
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Your solution:
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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):
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self-critique rating #$&*:
Very good work on the first four problems. Just one or two glitches there.
I've inserted Given Solutions for comparison.
Be sure to use updated versions of the Queries. I'll be posting more given solutions on Tuesday.
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