Application of partial derivatives

  • Thread starter WY
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  • #1
WY
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Hey,

I have no idea where to start, for this question. I know that I will probably have to use vector and scalar product and use the trig identity tan^2(theta)=sec^2(theta)-1 - and of course partial derivatives

Question:
In order to determine the angle theta which a sloping plane ceiling makes with the horizontal floor, an equilateral traingle of side-length l is drawn on the floor and the height of the ceiling above the three vertices is measured to be a,b and c. Show that:
tan^2(theta) = 4(a^2 + b^2 + c^2 - ab -bc - ac)/3t^2

Thanks in advance for anyone's help!
 

Answers and Replies

  • #2
saltydog
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WY said:
Hey,

I have no idea where to start, for this question. I know that I will probably have to use vector and scalar product and use the trig identity tan^2(theta)=sec^2(theta)-1 - and of course partial derivatives

Question:
In order to determine the angle theta which a sloping plane ceiling makes with the horizontal floor, an equilateral traingle of side-length l is drawn on the floor and the height of the ceiling above the three vertices is measured to be a,b and c. Show that:
tan^2(theta) = 4(a^2 + b^2 + c^2 - ab -bc - ac)/3t^2

Thanks in advance for anyone's help!
WY, I've seen your post elsewhere. Others may be like me: I don't know. However, if it were my problem and I HAD to solve it, I tell you what, you can be sure I'd be making a nice size cardboard, wood, plastic whatever sloping ceiling, nothing big, few feet maybe, draw my triangle on the floor, measure the distances and just start working with it. It's an interesting problem and I think, if no one helps you, that's a good place to start. :smile:
 
  • #3
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Solution

There is no need of partial derivatives. Here is the solution:

Consider the floor to be the x-y plane and the 3 vertices of the equilateral triangle A(0,0,0), B(t,0,0), C(t/2,(sq. root of 3)t/2,0).

The points in the plane of the ceiling will be
P(0,0,a), Q(t,0,b), R(t/2,(sq. root of 3)t/2,c)

The eqn of the plane of ceiling through these points (after simplificaton) is obtained as
(sq root 3)(a-b)x+(a+b-2c)y+(sq root 3)t(z-a) = 0
Eqn of x-y plane is z = 0

The angle theta between these planes is obtained as
cos(theta) = (sq root 3)t/(sq root of (3(a-b)^2+(a+b-2c)^2+3t^2) )

Using tan^2(theta) = sec^2(theta)-1 and symplifying we get
tan^2(theta) = 4(a^2+b^2+c^2-ab-bc-ca)/3t^2
 

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