# Hard arc length problem

## Homework Statement

A hawk flying at 2 m/s at an altitude of 80 m accidentally drops its prey. The parabolic trajectory of the falling prey is described by the equation below until it hits the ground, where y is its height above the ground and x is its horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until the time it hits the ground. Express your answer correct to the nearest tenth of a meter.

y = 80 - x^2/80

ans : ____m

## The Attempt at a Solution

arrggggg: here it is :

y' = -1/40x

1 + (f '(x)) = 1 + 1/1600 x^2

$$\int\sqrt{1 + 1/1600*x^2}$$

=

1/40$$\int \sqrt{1600 + x^2}$$

using trig sub and letting x = 40tan( $$\Theta$$ )
=

40 $$\int sec( \Theta )^3$$

=

$$\frac{40}{41}$$$$sec(\vartheta)tan(\vartheta) + ln| sec(\vartheta)tan(\vartheta) |$$

so,

1/40$$\int \sqrt{1600 + x^2}$$ = $$\frac{40}{41}$$$$sec(\vartheta)tan(\vartheta) + ln| sec(\vartheta)tan(\vartheta) |$$

I am very lost if this is the integral. Any help please....

Double check antiderivative: missing a plus sign.

Not sure where 41 came from. Should be 2? Missing parentheses, too.

Now convert back to x.

the 41 is from the from the integral of 1/40 * sec^3 x

Well, what is the integral of sec^3 x by itself?

This is what I got substituting it :

20 ( sqrt(40+x^2) / x * x/40 + ln | sqrt(40+x^2) / x * x/40 | ) ( from 0 ->80 ?)

not sure what the limits are.

edited

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What is sec(theta)? What is tan(theta)? And where's that missing plus sign?

Limits look good.

sec(theta) = sqrt(40+x^2)
tan (theta) = x/40

since originally x = 40tan(theta)

Hello tnutty,

You are making things more difficult than necessary. The integral to solve is the following:
$$\int \sqrt{a^2+x^2}dx$$
By doing the substitution you proposed, you end up with
$$\int \frac{d \theta}{cos^3(\theta )}$$
This is not easier than the first one :-( What you can do is start with partial integration, giving:
$$I=x\sqrt{a^2+x^2}-\int \sqrt{a^2+x^2}dx+a^2\int \frac{dx}{\sqrt{a^2+x^2}}$$
One of these is equal ti I itself and the other one can be solved almost directly by setting $x=a sinh(t)$
It looks difficult, but it is a lot easier than the trig way you used. The problem there is that you don't have the basic goniometric relation $sin^2(x)+cos^2(x)=1$ in it, but you do have $$cosh^2(x)-sinh^2(x)=1[/itex] for the last integral. Try it and let us know if this works. coomast [tex] \int \frac{d \theta}{cos^3(x) dx}$$

ALL the integral below should have dx but it did not register.

$$\int cos^2(x) * cos(x)dx$$

cos^2(x) = 1-sin^2(x)

$$\int(1- sin^2(x) * cos(x)dx$$

u = sin(x)
du = cos(x) dx

$$\int( 1 - u^2) du$$

=

u - 1/3u^3

=

sin(x) - sin^3(x) / 3 + C

I think this might be a easier way to integrate than what was suggested, at least to me.

I have one question. How did you find that the integral to solve is :

$$\int \sqrt{a^2+x^2}dx$$

Hello tnutty,

Let's reframe this step by step. First the setting up of the integral is correct, you got:

1/40$$\int \sqrt{1600 + x^2}$$

Which is the same as the one I gave:

$$\int \sqrt{a^2+x^2}dx$$

I left out the factor before the integral and set $1600=a^2$ to make the writing a bit easier. So no discussion on this part so far. The next step is to solve the integral, you used a trig substitution and obtained (beware of the naming, you're teacher will certainly not be pleased on an exam if you do this):

$$\int \frac{d \theta}{cos^3(x) dx}$$

Now, why are you solving this as:

$$\int\ cos^3(\theta)d\theta$$

It is a different function. So the integral to solve has to be:

$$\int\ \frac{1}{cos^3(\theta)}d\theta$$

This is where I would try to find something else and that was what I explained in my earlier post.

Hope this helps,

coomast

With your original approach, I think of $$\int\sec^3\theta\,d\theta$$ as a standard example. It will be in any table, and it will be almost certainly be worked out as a sample problem in your calc book. Look it up and you'll see you almost have it (but you're missing a plus sign).

The problem is that substituting back to get in terms of x, your tan(theta) is correct, but your sec(theta) isn't. If you drew a triangle, you labeled it wrong or are using it wrong. If you used a trig identity, it's not right, or you used it wrong.

Or use coomast's approach.

Cyosis
Homework Helper
It's not an easy integral if you don't know the hyperbolic functions, but your substitution will work as well it will just take a bit more work.

Here is a start.

\begin{align*} \int \sec^3 \theta\, d\theta &= \int \frac{1}{(1-\sin^2 \theta)^{\frac{3}{2}}}\, d\theta \\ &= \int \frac{1}{(1-u^2)^2}\, du \\ &= \int \left(\frac{1}{(1-u)(1+u)}\right)^2\, du \\ &= \frac{1}{4} \int \left(\frac{1}{(1-u)^2}+\frac{1}{(1+u)^2}+\frac{1}{1-u}+\frac{1}{1+u}\right)\, du \end{align*}

Try to check the steps and continue from here on. The substitution $u=\sin \theta$ was used.

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First IBP, then use trig identity:

\aligned I&=\int\sec^3\theta\,d\theta \\ &=\sec\theta\tan\theta-\int\sec\theta\tan^2\theta\,d\theta \\ &=\sec\theta\tan\theta-\int\sec\theta(\sec^2\theta-1)\,d\theta \\ &=\sec\theta\tan\theta-\int\sec^3\theta\,d\theta+\int\sec\theta\,d\theta \\ &=\sec\theta\tan\theta-I+\ln|\sec\theta+\tan\theta| \endaligned

Now add I to both sides and divide by 2:

$$I=\frac12\sec\theta\tan\theta+\frac12\ln|\sec\theta+\tan\theta|+C$$

Or memorize it! :)

Sweet, I figured out the answer = 118.3.

I was using the wrong trig identity. Should have used x = 40tan(theta).

If you want the solution let me know. Otherwise thanks for helping me.

Such a long process for this problem. Feel like shooting that hawk for dropping its food.

Sweet, I figured out the answer = 118.3.

I was using the wrong trig identity. Should have used x = 40tan(theta).

If you want the solution let me know. Otherwise thanks for helping me.

Such a long process for this problem. Feel like shooting that hawk for dropping its food.

Very good, you found it. It can be written as: $L=20\cdot [2\sqrt{5} + ln(2+\sqrt{5})]$

best regards,

coomast