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hyperbolic Functions

The curve assumed by a heavy power cable is described by the equation...


y= 60 cosh (x/60) where x and y are horizontal and vertical positions respectively.

Calculate the value of X When Y is 180

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Original post by HNCSTUDENT1
The curve assumed by a heavy power cable is described by the equation...


y= 60 cosh (x/60) where x and y are horizontal and vertical positions respectively.

Calculate the value of X When Y is 180


So what's preventing you from just solving 180=60cosh(x60)\displaystyle 180=60 \cosh \left( \frac{x}{60} \right)??
1. Substitute y for 180
180 = 60 cosh (x/60)

2. Take the cos60 to one side to get the x function on the other side
180/cosh60 = x/60

3. Take the 60 over to the other side to find x
60(180/cosh60) = x

Hope that answers your question :smile:
(edited 6 years ago)
Original post by iTzYoBoY

2. Take the cos60 to one side to get x on the other side
180/cos60 = x/60


????
Original post by iTzYoBoY
1. Substitute y for 180
180 = 60 cos (x/60)

2. Take the cos60 to one side to get x on the other side
180/cos60 = x/60

3. Take the 60 over to the other side to find x
60(180/cos60) = x

Hope that answers your question :smile:


it's cosh, not cos.

:spank:
Original post by iTzYoBoY
1. Substitute y for 180
180 = 60 cos (x/60)

2. Take the cos60 to one side to get the x function on the other side
180/cos60 = x/60

3. Take the 60 over to the other side to find x
60(180/cos60) = x

Hope that answers your question :smile:


The function in the question is cosh (...) which is the hyperbolic cosine function, not cos.

Having said that, your working out doesn't
make sense unfortunately.
Reply 6
Original post by RDKGames
So what's preventing you from just solving 180=60cosh(x60)\displaystyle 180=60 \cosh \left( \frac{x}{60} \right)??


it doesn't work
Original post by HNCSTUDENT1
it doesn't work


What do you mean it doesn't work? cosh(x)160cosh(x)60\cosh(x) \geq 1 \Rightarrow 60\cosh(x) \geq 60 so y=180y=180 should work just fine...

If you get stuck in your working, post it here.
Reply 8
Original post by RDKGames
What do you mean it doesn't work? cosh(x)160cosh(x)60\cosh(x) \geq 1 \Rightarrow 60\cosh(x) \geq 60 so y=180y=180 should work just fine...

If you get stuck in your working, post it here.


How does this allow me to find x?
Reply 9
Original post by HNCSTUDENT1
The curve assumed by a heavy power cable is described by the equation...


y= 60 cosh (x/60) where x and y are horizontal and vertical positions respectively.

Calculate the value of X When Y is 180


Obtaining x is a similar method to trigonometric functions.

Can you re-arrange so that you have just on one side:

cosh(x/60) \cosh(x/60) ?
Original post by HNCSTUDENT1
How does this allow me to find x?


It doesn't, I'm just showing to you that it does work because there are valid solutions to the equation.

Anyway, if you have attempted the question, post your attempt here.
Did you get your head around this in the end, I'm having difficulty with the same question
Original post by gavalar2010
Did you get your head around this in the end, I'm having difficulty with the same question


So, what have you done so far? Any thoughts?
Using the definition of cosh in terms of ex/60 will lead to solving a quadratic with a positive discriminant.
Why would it not work?
I need a point in the right direction on getting started on this.
Original post by gavalar2010
I need a point in the right direction on getting started on this.

Er...

Original post by dextrous63
Using the definition of cosh in terms of ex/60 will lead to solving a quadratic with a positive discriminant.
Why would it not work?
Ok thanks. It may as well be written in hieroglyphics to me at the moment. 👍
Reply 17
Original post by gavalar2010
Ok thanks. It may as well be written in hieroglyphics to me at the moment. 👍

Essentially we dealing with equation:
180=60cosh(x/60)=60(ex/60+ex/602). \displaystyle 180 = 60 \mathrm{cosh} (x/60) = 60 \Big( \frac{e^{x/60} + e^{-x/60}}{2} \Big).

Are you able to rearrange so to have cosh \mathrm{cosh} (or even better: ex/60 e^{x/60} ) as the subject?
Reply 18
180 = 60cosh(x/60)
3 = cosh(x/60)
6 = e^(x/60) + e^-(x/60)
0 = e^(2x/60) - 6e^(x/60) + 1
Find e^(x/60) (ANS)
x = +- 60ln(ANS)
so is X= 60(180/cosh60) incorrect?

thats how i rearranged it prior to reading this thread

=1.8914063e-22
(edited 4 years ago)

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