Calculating Distance Traveled (Arc Length)

[colonelone]

Homework Statement

Given y = 156 - (x - 40)²/60. x = 0 and x = 85 Find distance traveled

Homework Equations

Arc Length S = integral of square root of ( y' )²

The Attempt at a Solution

Doing this I get the trig sub tan(t) = (x - 40)/ 30 (Told by teacher to use this instead of -(x - 40)/30 )
so x = 30tan(t) + 40
and dx = 30sec²(t)

So then under the radical I get 1 + ( - (tan (t) )²
So is this equal to the square root of sec²(t)?

Assuming it is, I get the integral of sec²(t) times 30sec²(t) or sec³(t)

I know how to do this (I think) but I'm really not sure (I can't use what's written in the back of my book).

Doing the work I get:

S(x) = 15sec(t)tan(t) + (1/2) ln[ (sec(t) + tan(t) ]
Substitution x back in for t gives

tan(t) = (x-40)/30
and
sec(t) = square root [ 1 + ( ( -x + 40)/(30) )²

But evaluating this from 0 to 85 gives me an arc length of ~ 75.05, which I know can't be right, but I can't figure out where I went wrong.

 

[cronxeh]

Isn't it S= integral (sqrt( 1+ (y')^2)) ?

y' = (40-x)/30
(y')^2 = ((40-x)^2)/900
integral sqrt(1+((40-x)^2)/900), x=0..85 ~ 108.296

As a matter of fact you could've estimated this distance by taking mid-way point (in this case its an inflection point (x-40)/30 = 0 -> x=40

at x=40, y=156
at x=0, y=129.3
at x=85, y=122.25

As you can see you can draw 2 lines from x=0 to x=40, and from x=40 to x=85, then add up the distances to get your approximate arc length:

sqrt((0-40)^2+(156-129.3)^2) + sqrt((85-40)^2 + (156-122.25)^2) ~ 104.34

 

[colonelone]

Isn't it S= integral (sqrt( 1+ (y')^2)) ?

y' = (40-x)/30
(y')^2 = ((40-x)^2)/900
integral sqrt(1+((40-x)^2)/900), x=0..85 ~ 108.296

As a matter of fact you could've estimated this distance by taking mid-way point (in this case its an inflection point (x-40)/30 = 0 -> x=40

at x=40, y=156
at x=0, y=129.3
at x=85, y=122.25

As you can see you can draw 2 lines from x=0 to x=40, and from x=40 to x=85, then add up the distances to get your approximate arc length:

sqrt((0-40)^2+(156-129.3)^2) + sqrt((85-40)^2 + (156-122.25)^2) ~ 104.34

What equation did you use to evaluate s(x)?

Once I redid/rechecked everything I got

s(x) = 15 [ sqrt [ 1 + ( ( -x + 40)/(30 ) )² ] * ( ( x - 40 ) / (30) ) + ln [ sqrt [ 1 + ( ( -x + 40)/(30 ) )² + ( ( x - 40 ) / (30) ) ]

 

[cronxeh]

$$S = \int^{85}_{0} \sqrt{1+\frac{(40-x)^{2}}{900}} dx$$

 

[colonelone]

$$S = \int^{85}_{0} \sqrt{1+\frac{(40-x)^{2}}{900}} dx$$

I mean after all the trig sub, the final equation that you substitute 85 and 0 into.

 

[cronxeh]

I mean after all the trig sub, the final equation that you substitute 85 and 0 into.

1/60 sqrt((x-40)^2+900) (x-40)+15 sinh^(-1)((x-40)/30)

 

[colonelone]

1/60 sqrt((x-40)^2+900) (x-40)+15 sinh^(-1)((x-40)/30)

We haven't learned about hyperbolic functions yet. Is there a way to do it without using sinh, i.e. by using trig sub and integration by parts.

 

[cronxeh]

We haven't learned about hyperbolic functions yet. Is there a way to do it without using sinh, i.e. by using trig sub and integration by parts.

Sure, its just what you did but with the correct limits, which I think you should double check

Let u=40-x, du=-dx, now do the limits here

integral -sqrt(1+(u^2)/900) du

then substitute u=30*tan(s), du=30*sec^2(s)ds

= integral -sqrt(tan^2(s)+1) =
where s=arcan(u/30)

integral -30*sec^3(s)ds =
-15*tan(s)sec(s) - 15*integral(sec(s)ds)
= -15tan(s)sec(s) - 15*ln(tan(s)+sec(s))
sub back s=arctan(u/30) and u=40-x:

1/60*(x-40)*sqrt(x^2-80*x+2500) - 15*ln(1/30*(sqrt(x^2-80*x+2500)-x+40))
=58.4839-(-49.8125) = 108.2964

 

[colonelone]

Sure, its just what you did but with the correct limits, which I think you should double check

Let u=40-x, du=-dx, now do the limits here

integral -sqrt(1+(u^2)/900) du

then substitute u=30*tan(s), du=30*sec^2(s)ds

= integral -sqrt(tan^2(s)+1) =
where s=arcan(u/30)

integral -30*sec^3(s)ds =
-15*tan(s)sec(s) - 15*integral(sec(s)ds)
= -15tan(s)sec(s) - 15*ln(tan(s)+sec(s))
sub back s=arctan(u/30) and u=40-x:

1/60*(x-40)*sqrt(x^2-80*x+2500) - 15*ln(1/30*(sqrt(x^2-80*x+2500)-x+40))
=58.4839-(-49.8125) = 108.2964

For some reason when I plug in the numbers I get an arc length of 118.693625.

Also, where did the 1/60 and 1/30 come from? It seems like you were plugging in ( 40 - x) / 900 for tan(s) instead of ( 40 - x ) /30

 

[cronxeh]

For some reason when I plug in the numbers I get an arc length of 118.693625.

Also, where did the 1/60 and 1/30 come from? It seems like you were plugging in ( 40 - x) / 900 for tan(s) instead of ( 40 - x ) /30

= -15tan(s)sec(s) - 15*ln(tan(s)+sec(s))
sub back s=arctan(u/30) and u=40-x:

If you agree with that line, then watch the magic:

tan(arctan(x)) = x
cos(arctan(x)) = 1/sqrt(x^2+1)
sec(arctan(x)) = sqrt(x^2+1)
s=arctan((40-x)/30)

-15tan(s)sec(s) - 15*ln(tan(s)+sec(s)) =
-15*ln(tan(s) + 1/cos(s)) - (15*tan(s))/cos(s) =
-15*((40-x)/30)*sqrt(((40-x)/30)^2+1) - 15*ln(((40-x)/30)+sqrt(((40-x)/30)^2+1)) =
(x/2 - 20)*((x/30 - 4/3)^2 + 1)^(1/2) - 15*ln(((x/30 - 4/3)^2 + 1)^(1/2) - x/30 + 4/3)
Plug in your limits x=85, x=0:

58.4839 - (-49.8125) = 108.2964

If you are not a believer in the answer, consider that we forget calculus for a second and just approximate it:

Given y = 156-(x-40)^2/60, let's do 10 intervals from x=0 to x=85:
distance traveled is then ( in MATLAB ):

x=0:85
d=0;
for i=1:85
d=d+sqrt(((156-(x(i)-40)^2/60) - (156-(x(i+1)-40)^2/60))^2 + (x(i)-x(i+1))^2);
end
d

d =

108.2942

If you want me to go more accurate,

x=0:0.1:85
d=0;
for i=1:850
d=d+sqrt(((156-(x(i)-40)^2/60) - (156-(x(i+1)-40)^2/60))^2 + (x(i)-x(i+1))^2);
end
d

d =

108.2964