Allowable error in elevation angle in pyramid of giza

[minia2353]

Homework Statement

the base of the great pyramid giza is a square that is 230m on each side.
use differentials to estimate the allowable error in the elevation angle that will ensure that the error in calculating the height is at most +,- 5m.
phi is the elevation angle of the pyramid.
h= height of the pyramid.

Homework Equations

The Attempt at a Solution

I've tried :

h= 115 tan (phi)
differentiate h with respect to phi = 115 sec^2 (phi)
dh/h= 115sec^2 d phi/ 115tan (phi)

and I don't know what to do next .

 

[Mark44]

Homework Statement

the base of the great pyramid giza is a square that is 230m on each side.
use differentials to estimate the allowable error in the elevation angle that will ensure that the error in calculating the height is at most +,- 5m.
phi is the elevation angle of the pyramid.
h= height of the pyramid.

Homework Equations

The Attempt at a Solution

I've tried :

h= 115 tan (phi)
differentiate h with respect to phi = 115 sec^2 (phi)
dh/h= 115sec^2 d phi/ 115tan (phi)

and I don't know what to do next .

This is a pretty good start, although you have a mistake in your last line. It should be 115 sec^2(phi) d phi in the numerator.

I'm going to shorten phi to p, since my keyboard doesn't have Greek letters on it. Hopefully you will be able to translate my p's back to phi's.

Backing up a little, you have dh = 115 sec^2(p) dp. What you need is an equation that is solved for dp in terms of dh and h, which is assumed to be some constant that you don't happen to know.

Here's what you have to work with:
|dh| <= 5 m.
A right triangle whose legs are 115 and h (meters).

Can you continue from here?

 

[minia2353]

I know that I need to get that equation and I've tried but I don't seem to get it.
That's where I need help.

 

[Mark44]

1. Solve dh = 115 sec^2(p) dp for dp.
2. Use the right triangle whose legs are 115 and h, with acute angle p to solve for p in terms of h.
3. Substitute for p in the equation from step 1.

 

[minia2353]

It's the 2nd part of your step I'm stuck on.

 

[Mark44]

You mean step 2? Solve for p in the right triangle in the attached image.

 

[minia2353]

Yes,
I konw the triangle but I still can't get at it.
It seems like a simple one but I don't seem to know something to get at it.

 

[Mark44]

After you figure out the hypotenuse you can write an equation with tan(p), sin(p), or cos(p). To solve for p, use the appropriate inverse function.

 

[minia2353]

does it become this?
dp= dh/115sec^2tan^-1 (h/115)

 

[Mark44]

Well, sort of, but it can be greatly simplified.

dh = 115sec²(p)dp ==> dp = (1/115)(cos²(p)*dh)

You have replaced p by tan^-1(h/115). This isn't wrong, but a better choice would be to write p in terms of the cos^-1 function. If you do that, it will make it easier to get bounds on dp.

 

[minia2353]

Then would it be this ; simplified + cos^-1
:
dp=( 1/115 ) (cos^2 (115/(√h^2 + 115^2)dh)

 

[Mark44]

Nope, try again. cos should be completely gone, and there is another error.

 

[minia2353]

I forgot to place cos^-1 in front of 115/√h^2 + 115^2
and the 115^2 goes under the sign with h^2 as well.

 

[Mark44]

Yes, but there is another error in 115/√h^2 + 115^2.

Also, since there are two terms in the denominator, put parentheses around the whole denominator.

 

[minia2353]

another error??

 

[Mark44]

Yes. Your value for cos(p) is wrong. Or maybe it's the way you have written it: 115/√h^2 + 115^2.

What this means to me is
\frac{115}{(\sqrt{h})^2} + 115^2
which is equal to
\frac{115}{h} + 115^2

If this is NOT what you meant, then you need to write this as 115/√(h^2 + 115^2) or 115/sqrt(h^2 + 115^2).

 

[minia2353]

The equation you understood it as is not what I meant,
I'm sorry for the error.
It's 115/ sqrt (h^2+ 115^2)

 

[minia2353]

the equation is up to now is:
dp= (1/115) (cos^2 cos^-1 (115/ sqrt (h^2+ 115^2) ) dh)

 

[Mark44]

Yes, but this expression can still be simplified.

cos^2(cos^{-1}(a)) = [cos(cos^{-1}(a)]^2 = ?