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Calc 3 11.6

 

<== Calculus 3
Number Question Answer
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1. Find the directional derivative of f(x,y) = x^2 y^3 + 2x^4 y at the point (-3, 2) in the direction 	heta = \pi/3.
The gradient of f is:
&nabla f = &lang , &rang
&nabla f (-3, 2) = &lang, &rang
The directional derivative is:
&nabla f =
&lang,&rang
&nabla f (-3, 2) = &lang -480 , 270 &rang
directional derivative = -6.173
The work
2. Consider the function f(x,y,z) = xy + yz^2 + xz^3. Find the gradient of f:
&lang, , &rang
Find the gradient of f at the point (-3, -1, -3).
&lang, , &rang
Find the rate of change of the function f at the point (-3, -1,-3) in the direction \mathbf u = \langle 3/\sqrt{14}, 1/\sqrt{14}, -2/\sqrt{14} 
angle.
&nabla f =
&lang&rang
&nabla f (-3, -1, -3) = &lang -28, 6, -75&rang
rate of change = 19.24
The work
3. Find the directional derivative of f(x,y,z) = z^3 - x^2 y at the point (-3, -3, -5) in the direction of the vector
v = &lang -1, -5, -1 &rang .
-2.31
The work
4. Find the directional derivative of f(x,y,z) = 5 x^2  - 2 y^2  - 5
z^2 at the point (2, 5, 3) in the direction of the origin.
24.33
The work
5. Find the maximum rate of change of f(x,y,z) = x + y/z at the point (2, 2, 2) and the direction in which it occurs.
Maximum rate of change:
Direction (unit vector) in which it occurs:
&lang, ,&rang
Max rate change: 1.225
Direction Vector:
&lang 0.8165, 0.4082, -0.40825&rang
The work
6. Find the maximum rate of change of f(x,y) = \ln(x^2 + y^2) at the point (3, 5) and the direction in which it occurs.
Maximum rate of change:
Direction (unit vector) in which it occurs: &lang, &rang
Maximum rate change: 0.343
Direction Vector:
&lang 0.5145, 0.8575&rang
The work
7. Consider the function f \left( x, y 
ight) = 2x^2 + 4y^2.
Find the the directional derivative of f at the point ( 3, -2 ) in the direction given by the angle 	heta = rac{2\pi}{3}.



Find the unit vector which describes the direction in which f is increasing most rapidly at ( 3, -2 ).
(, ).

Directional derivative: -19.856

Unit vector: ( 0.6, -0.8 )
The work
8. Suppose f (x,y) = x/y, P = (-1, 2) and v = -4i - 4j.

A. Find the gradient of f.
&nabla f = i + j

B. Find the gradient of f at the point P.
(&nabla f) (P) = i + j

C. Find the directional derivative of f at P in the direction of v.
D_{u} f =

D. Find the maximum rate of change of f at P.

E. Find the (unit) direction vector in which the maximum rate of change occurs at P.
u = i + j

A) (1/y)i + (-x/y^2)j

B) .5i + .25j

C) -0.5303

D) 0.559

E) 0.894i + 0.447j
The work
9. Suppose that you are climbing a hill whose shape is given by z = 692 - 0.08 x^2 -0.1 y^2, and that you are at the point (20, 60, 300).
In which direction (unit vector) should you proceed initially in order to reach the top of the hill fastest?
&lang,&rang
If you climb in that direction, at what angle above the horizontal will you be climbing initially (radian measure)?
Proceed: &lang -0.2577, -0.9662&rang

1.4905 rad
The work
10. Consider a function f (x,y) at the point (6, 3).
At that point the function has directional derivatives:
rac{6}{\sqrt{58}} in the direction (parallel to) &lang 7, 3 &rang, and
rac{6}{\sqrt{52}} in the direction (parallel to) &lang 6, 4 &rang . The gradient of f at the point (6, 3 ) is
(, ).
( 0.6, 0.6)
The work
11. Find equations of the tangent plane and normal line to the surface x = 2 y^2 + 4 z^2  - 593 at the point (7, -10, 10).
Tangent Plane: (make the coefficient of x equal to 1).
= 0.
Normal line: &lang 7, , &rang
+ t&lang 1, , &rang .
Tangent Plane: x+40y-80z+1193=0

Normal line:
&lang 7, -10, 10 &rang + t&lang 1, 40, -80&rang
The work
12. Find equations of the tangent plane and normal line to the surface x = 2 y^2 + 4 z^2  - 593 at the point (-9, 0, 0).
Tangent Plane: (make the coefficient of x equal to 1).
= 0.
Normal line: &lang -9, , &rang
+ t&lang , , 1 &rang .
Tangent Plane: -x + 9y + z -9

Normal line:
&lang -9, 0, 0 &rang + t&lang -1, 9, -1&rang
The work
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