Calc problem: Help please
A cable hangs between two poles of equal height and 33 feet apart. At a point on the ground directly under the cable and x feet from the point on the ground halfway between the poles
the height of the cable in feet is
h(x)=10 +(0.3)( x^{1.5}) .
The cable weighs 12.3 pounds per linear foot.
Find the weight of the cable.

Question

Calc problem: Help please
A cable hangs between two poles of equal height and 33 feet apart. At a point on the ground directly under the cable and x feet from the point on the ground halfway between the poles
the height of the cable in feet is 
h(x)=10 +(0.3)( x^{1.5}) .
The cable weighs 12.3 pounds per linear foot.
Find the weight of the cable.

anonymous · Answer

If you put the function h(x) into the arc length formula you get its length, then multiply by 12.3 to get the total weight.

anonymous · Answer

$$L= \int\limits_{a}^{b} \sqrt{1+(10+(0.3)(x ^{1.5})} *(12.3)$$

anonymous · Answer

what would the value of x be?

anonymous · Answer

under the radical it should be 1+(h'(x))^2 not 1+h(x) so find h'(x) and square it. integrate with respect to x and use a=0 and b=33 feet. Then multiply by 12.3

anonymous · Answer

$$L= \int\limits_{0}^{33} \sqrt{1+(\frac{ 9\sqrt{x} }{ 20 }})^2$$

*then multiply by 12.3

i got 821.763
@slickwilly

anonymous · Answer

I'm not sure about that integral. These ones are usually hard to integrate.
 Maybe someone else will figure it out but I can't and I'm pretty sure 821 ft is way too big.

anonymous · Answer

help? @campbell_st (:

campbell_st · Answer

sorry can't offer anything

anonymous · Answer

Thanks for looking into the problem though!

anonymous · Answer

I still got a very large answer... @campbell_st

anonymous · Answer

I just noticed h(x) goes from the midpoint to one of the ends so I think you need to integrate from 0 to 33/2 and then double the result.

anonymous · Answer

I still got a large answer in the thousands.

campbell_st · Answer

ok... just had a quick read

you need $$h'(x) = 0.45x^{0.5}$$

then the parabolic arc length is 

$$L = \int\limits_{a}^{b} \sqrt{1 + (h'(x))^2} dx$$

which means you are looking at 

$$L = \int\limits_{0}^{33} \sqrt{1 + 0.45^2x} dx... or L = \int\limits_{0}^{33}\sqrt{1 + 0.2025x} dx$$

anonymous · Answer

would I just evaluate both integrals and guess/check?

anonymous · Answer

do I have to multiply L by 12.3

anonymous · Answer

The last integral looks right but do it from 0 to 33/2 and then double it. Then multiply by 12.3

anonymous · Answer

okay I got 19.04 doubling it I got 38.08 then multiplying it by 12.3 I got 468.384 and that was wrong... ;/ @slickwilly

anonymous · Answer

I think I'm out of ideas for this one. Good luck.