gradient vectors: The axis of a light in a lighthouse is ...

[peacefreak77]

The axis of a light in a lighthouse is tilted. When the light points east, it is inclined upward at 4 degree(s). When it points north, it is inclined upward at 5 degree(s). What is its maximum angle of elevation?

I have no idea what to do. We are currently learning gradient vectors to a point on a 3D surface. I'm not sure where to start with this. Mostly I have been asked what the maximum gradient vector is (magnitude of gradient vector). I'm not sure how the angle of elevation can apply to all of that.

 

[JakeD]

Re: gradient vectors: The axis of a light in a lighthouse is

The axis of a light in a lighthouse is tilted. When the light points east, it is inclined upward at 4 degree(s). When it points north, it is inclined upward at 5 degree(s). What is its maximum angle of elevation?

I have no idea what to do. We are currently learning gradient vectors to a point on a 3D surface. I'm not sure where to start with this. Mostly I have been asked what the maximum gradient vector is (magnitude of gradient vector). I'm not sure how the angle of elevation can apply to all of that.

Here is how I solved this. It may not fit with what you've learned, but maybe it will give you some ideas.

First I determined the equation of the tilted plane through the origin the light is on. Define the x-axis as east-west, the y-axis as north-south, the z-axis as up-down. The equation of the tilted plane is ax+by+z=0. There are two points to determine the coefficients a and b of that plane: (-cos(t1),0,sin(t1)) when the light points east, and (0,cos(t2),sin(t2)) when the light points north. Then a = tan(t1), b = -tan(t2).

Now consider the intersection of the sphere with radius 1 and the tilted plane. The maximum elevation will be where the z coordinate is maximized on that intersection. So maximize z subject to the constraints ax+by+z=0 and x^2+y^2+z^2=1. Using Lagrange multipliers I found the solution to be z* = sqrt(1-lambda) where lambda = 1/(a^2+b^2+1) is the multiplier on the ax+by+z=0 constraint. The maximum angle of elevation is arcsin(z*) = 6.39 degrees.

 

[peacefreak77]

Okay, I know your answer is correct as my online homework program told me it was, now I need to understand how you got it.

I understand the dimensions of the tangent plane, but I don't understand how you got the points. (A and B are the slopes in the x and y direction, right, or the differential?). What is t? Are the points the origin of the light in the lighthouse or what? What angle are you using trig on?

With the sphere, are you saying that the maximum z coordinate is just the maximum z value within the cross-sectional circle the intersection of the sphere and the plane creates? And why use a sphere, or with radius 1?

Thank you for helping me.

 

[JakeD]

Okay, I know your answer is correct as my online homework program told me it was, now I need to understand how you got it.

I understand the dimensions of the tangent plane, but I don't understand how you got the points. (A and B are the slopes in the x and y direction, right, or the differential?). What is t? Are the points the origin of the light in the lighthouse or what? What angle are you using trig on?

With the sphere, are you saying that the maximum z coordinate is just the maximum z value within the cross-sectional circle the intersection of the sphere and the plane creates? And why use a sphere, or with radius 1?

Thank you for helping me.

t1 and t2 are the given angles of elevation, 4 and 5 degrees. Picture the lighthouse being at the origin with the light sweeping around on the plane and intersecting the unit sphere. We use the unit sphere to be able to use basic trig to translate between the angle of elevation and direction of the light and the coordinates of points on the sphere.

The equation for the plane has coefficients a and b for x and y. All points (x,y,z) are constrained to be on the plane and on the sphere.

To see that for a point (x,y,z) on the sphere, \(\displaystyle t = \arcsin(z)\) is the angle of elevation, consider the two vectors \(\displaystyle v_1 = (x,y,z)\) and \(\displaystyle v_2 = (x,y,0)\) and use the formula for the angle between them \(\displaystyle cos(t) = v_1 \cdot v_2 / |v_1| |v_2|\) where the dot denotes dot or inner product and |.| means the length. You should derive that \(\displaystyle t = \arcsin(z)\). All you'll need from trig is that \(\displaystyle sin^2(t) + cos^2(t) = 1.\)

If you haven't been introduced to all this stuff, I don't know how you would solve this problem. There may be an easier way, but I don't know it.

 

[peacefreak77]

we have done everything you mentioned. i'm just not so good at applying it yet i guess.

thank you. that made sense.