7. At what point on the paraboloid y = x^2 +z^2 is the tangent plane parallel to

Question

7. At what point on the paraboloid y = x^2 +z^2 is the tangent plane parallel to the plane x+2y+3z = 1?

8. If f(x, y) = xy, find the gradient vector ?f(3, 2) and use it to find the tangent line to the level
curve f(x, y) = 6 at the point (3, 2). Sketch the level curve, the tangent line, and the gradient
vector.

9. Near a buoy, the depth of a lake at the point (x, y) is z = 200 + 0.02x^2 ? 0.001y^3, where the
coordinates x and y are measured in meters, as is the depth z. A fisherman in a small boat
starts at the point (80, 60) moves towards the buoy, which is located at (0, 0). Is the water
under the boat getting deeper or shallower when the fisherman departs? Explain your answer.

Explanation / Answer

7) To find the tangent plane to the surface f(x,y,z)=0 at a point (X,Y,Z) we use the following method:

Calculate grad f = (f_x, f_y, f_z). The normal vector to the surface at the point (X,Y,Z) is grad f(X,Y,Z). The equation of a plane with normal vector n which passes through the point p is (r-p).n=0, where r=(x,y,z) is the position vector. So the equation of the tangent plane to the surface through the point (X,Y,Z) is ((x,y,z)-(X,Y,Z)).grad f(X,Y,Z)=0.

Now in your case we have f(x,y,z)=y-x^2-z^2, so grad f=(-2x,1,-2z), and the equation of the tangent plane at the point (X,Y,Z) is

((x,y,z)-(X,Y,Z)).(-2X,1,-2Z)=0,

that is

-2X(x-X)+1(y-Y)-2Z(z-Z)=0,

i.e.

-2Xx+y-2Zz = -2X^2+Y-2Z^2. (1)

Now compare this equation with the plane

x + 2y + 3z = 1. (2)

The two planes a_1x+b_1y+c_1z=d_1, a_2x+b_2y+c_2z=d_2 are parallel when (a_1,b_1,c_1) is a multiple of (a_2,b_2,c_2). So the two planes (1),(2) are parallel when (-2X,1,-2Z) is a multiple of (1,2,3), and we have

(-2X,1,-2Z)=1/2(1,2,3)

for X=-1/4 and Z=-3/4. On the paraboloid the corresponding y coordinate is Y=X^2+Z^2=1^4+9^4=5/2.

So the tangent plane to the given paraboloid at the point (-1/4,5/2,-3/4) is parallel to the given plane.

8)grad(xy)*(x-3,y-2) = 0
<2,3>*(x-3,y-7) = 0
2*(x-3) + 3*(y-7) = 0
2x - 6 + 3y - 21 = 0
2x + 3y - 27 = 0
3y = -2x + 27
y = -7/3*(x) + 42/3

9) A question on directional derivatives.

This question can be written

"Find the directional derivative of z at the point (80, 60) in the direction of (0,0).

First off, find the directional derivative, which is the vector of partial derivatives

Partial derivative of z with respect to x is 0.04x, ad evaluating at the point where the boat starts off gives an x-component of 3.2.

Partial derivative of z with respect to y is -0.003y^2, and evaluating at the point where the boat starts off gives a y-component of -10.8)

So the directional derivative, or grad(z) = (3.2, -10.8).

Now, look at the vector from (80, 60) to (0, 0). This is the vector (-80, -60). You should be able to verify that its magnitude = 100, and dividing by this magnitude gives its unit vector equivalent, (-0.8, -0.6).

Take the dot product of this with grad(z). If you get a positive answer, the water is getting deeper. If you get a negative answer, it's getting shallower.

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