# Math Help - Problem solving, hight, minimum...

1. ## Problem solving, hight, minimum...

Hi to all here is the question and I'll be providing my own answer which I guess it's wrong could any one tell me where I did mistake please?
Q:
The new rubish removal trucks were too high and were touching power cables.
Cables are hanging between poles along the side of a street with the poles 40 meters apart. the hight of the wires above the ground is given by:
y= 2(e^x/20 + e^-x/20)
the trucks are 3.4 m high and can extend 1.3 more while removing bins.
Explain whether the cables are high enough for the rubbish trucks to operate without becoming entangled in overhead cables strung between the power poles.
Mention the minimum height the power poles must be and also any change made to the equation for the centenary to accommodate the height change.
I've added 3.4 to 1.3 to get the overall hight which is: 4.7 right?
and I've placed it instead of x on the equation!
y= 2(e^4.7/20 + e^-4.7/20)
which is 4.10 and it's wrong i guess! isn't?

Thanks

2. Originally Posted by User Name
Hi to all here is the question and I'll be providing my own answer which I guess it's wrong could any one tell me where I did mistake please?
Q:

I've added 3.4 to 1.3 to get the overall hight which is: 4.7 right?
and I've placed it instead of x on the equation!
y= 2(e^4.7/20 + e^-4.7/20)
which is 4.10 and it's wrong i guess! isn't?

Thanks
Almost you want to put the 4.7 in for the y value (why?)

$4.7=2e^{\frac{x}{20}}+2e^{-\frac{x}{20}}$

Do you think you can finish from here?

3. Originally Posted by TheEmptySet
Almost you want to put the 4.7 in for the y value (why?)

$4.7=2e^{\frac{x}{20}}+2e^{-\frac{x}{20}}$

Do you think you can finish from here?
Hi, thanks.
So we put the 4.7 instead of Y not x? may I ask why?
I'll see how it goes...
we multiply 2e^x/20 x 2e^-x/20 right?
and we get 4e ^-x^2/40 ?
so we have:
4.7=4e^-x^2/40
we take 4 to to other side
0.7=e^-x^2/40
ln(0.7)=-x^2/40
and we take 40 other side:
40*ln(0.7)= -x^2
-14.26=-x^2
we have 3.77 which is wrong i guess where again i did mistake?

4. $4.7=2e^{\frac{x}{20}}+2e^{-\frac{x}{20}}
$

Multiply by $e^{\frac{x}{20}}$ to get

$4.7e^{\frac{x}{20}}=2\left(e^{\frac{x}{20}}\right) ^2+2
$

To simplify things let $u=e^{\frac{x}{20}}$ subbing this in we get

$4.7u=2u^2+2 \iff 2u^2-4.7u+2=0$

Now from here solve using the quadratic fomula.

Happy hunting

5. wow!
So you reckon will this give me the right answer? I'll give it try =}
Thanks.
mate, they way i went through was wrong ay?

6. here I got:
X=0.5580, X=1.7919
now we got the Xs do we have to plug them in?

7. Originally Posted by User Name
here I got:
X=0.5580, X=1.7919
now we got the Xs do we have to plug them in?

You are getting close

Remember we have solved for u and we need x

$e^{\frac{x}{20}}=1.79 \iff \frac{x}{20}=\ln(1.79) \iff x=20\ln(1.79) \approx 11.6$

So this tells us the x coordinate of where the truck will hit the wire.

So they are too low

The lowest point of the wire happen in the middle. I.e when x=20

8. Thanks I'll go trough the Q again.

9. OH god!!!
hey mate i went through them again and I don't get it =(
especially this part:
"So this tells us the x coordinate of where the truck will hit the wire.

So they are too low"
how the 11.6 can help us to do the question?

10. Originally Posted by User Name
OH god!!!
hey mate i went through them again and I don't get it =(
especially this part:
"So this tells us the x coordinate of where the truck will hit the wire.

So they are too low"
how the 11.6 can help us to do the question?
Here is a graph of the wire

I mispoke before the low point is at x=0

The point where the truck ran into the wire is correct.

So we need to shift the graph up 0.7 meters

So the new function will look like this

$y=2(e^{\frac{x}{20}}+e^{-\frac{x}{20}})+0.7$

The height of the poles will be

$y(20)=2(e^1+e^{-1})+0.7$