Calculating Distance Traveled (Arc Length)

• colonelone
In summary, the conversation discusses the process of finding the distance traveled given a mathematical equation and specific values for x. The process involves using trigonometric substitution and integration to find the arc length, and calculating the distance traveled based on this value. There is also a mention of using hyperbolic functions, but it is determined that the calculation can be done without using them. The conversation ends with a demonstration of approximating the distance traveled using MATLAB.
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.

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

Last edited:
cronxeh said:
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) ) ]

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

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

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

colonelone said:
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)

cronxeh said:
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.

colonelone said:
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

cronxeh said:
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

colonelone said:
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 ):
Code:
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,

Code:
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

Last edited:

What is arc length?

Arc length is the distance along the curved line of an arc. It is calculated by taking the length of the arc and multiplying it by the radius of the circle.

How do you calculate arc length?

To calculate arc length, you can use the formula L = rθ, where L is the arc length, r is the radius of the circle, and θ is the central angle in radians. You can also use the formula L = 2πr(n/360), where n is the central angle in degrees.

What is the unit of measurement for arc length?

Arc length is typically measured in units of length, such as centimeters, meters, or kilometers.

How does arc length differ from circumference?

Arc length is the distance along a curved line, while circumference is the distance around the outside of a circle. Arc length is a portion of the circumference.

Can arc length be calculated for any shape?

Arc length can only be calculated for circular shapes. For other shapes, such as squares or triangles, you would need to use different formulas to calculate the distance traveled.

Replies
2
Views
1K
Replies
7
Views
2K
Replies
1
Views
1K
Replies
7
Views
954
Replies
1
Views
2K
Replies
6
Views
1K
Replies
6
Views
2K
Replies
9
Views
2K
Replies
5
Views
2K
Replies
8
Views
2K