What went wrong when applying the FTOC to maximize arc length?

In summary, the arc-length is given by: $\displaystyle s=\int_0^{\frac{v_0^2}{g}\sin(2\theta)} \sqrt{1+\left(\frac{dy}{dx} \right)^2}\,dx$ where the maximum occurs for $\theta=45^{\circ}$.
  • #1
MarkFL
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Yesterday on another forum, someone asked for the launch angle which will maximize the arc-length of the trajectory for a projectile, assuming gravity is the only force on the projectile after the launch.

I eliminated the parameter t to get the trajectory:

$\displaystyle y=\tan(\theta)x-\frac{g\sec^2(\theta)}{2v_0^2}x^2$

and so the arc-length is given by:

$\displaystyle s=\int_0^{\frac{v_0^2}{g}\sin(2\theta)} \sqrt{1+\left(\frac{dy}{dx} \right)^2}\,dx$

Now, at this point I wanted to differentiate this using the derivative form of the FTOC, and what I got was:

$\displaystyle \frac{ds}{d\theta}=\frac{2v_0^2}{g}\cos(2\theta) \sqrt{1+\left(\frac{dy}{dx} \right)^2}$

This implies that the maximum occurs for:

$\displaystyle \theta=45^{\circ}$

Which I knew was incorrect. The OP stated the solution was approximately:

$\displaystyle \theta=56.5^{\circ}$

I gave up this route and went ahead and integrated through a myriad of substitutions to finally find:

$\displaystyle s(\theta)=\frac{v_0^2}{g}(\sin(\theta)+\cos^2( \theta)\ln(\sec(\theta)+\tan(\theta)))$

Differentiating and equating to zero eventually led to:

$\displaystyle f(\theta)=\sin(\theta)\ln(\sec(\theta)+\tan(\theta))-1=0$

and using Newton's method, I found:

$\displaystyle \theta\approx56.4658351275^{\circ}$

My question is, what did I do wrong when trying to apply the FTOC? I suspect I am not correctly applying the chain rule.
 
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  • #2
MarkFL said:
Yesterday on another forum, someone asked for the launch angle which will maximize the arc-length of the trajectory for a projectile, assuming gravity is the only force on the projectile after the launch.

I eliminated the parameter t to get the trajectory:

$\displaystyle y=\tan(\theta)x-\frac{g\sec^2(\theta)}{2v_0^2}x^2$

and so the arc-length is given by:

$\displaystyle s=\int_0^{\frac{v_0^2}{g}\sin(2\theta)} \sqrt{1+\left(\frac{dy}{dx} \right)^2}\,dx$

Now, at this point I wanted to differentiate this using the derivative form of the FTOC, and what I got was:

$\displaystyle \frac{ds}{d\theta}=\frac{2v_0^2}{g}\cos(2\theta) \sqrt{1+\left(\frac{dy}{dx} \right)^2}$

This implies that the maximum occurs for:

$\displaystyle \theta=45^{\circ}$

Which I knew was incorrect. The OP stated the solution was approximately:

$\displaystyle \theta=56.5^{\circ}$

My question is, what did I do wrong when trying to apply the FTOC? I suspect I am not correctly applying the chain rule.

You need to account for the fact that the integrand is also dependent on \(\theta\).

If I have done this right the FTOC you need is:

If:

\[s(\theta)=\int_a^{f(\theta)} h(x,\theta) dx\]

Then:

\[s'(\theta)=f'(\theta)h(f(\theta),\theta)+\int_a^{f(\theta)}\frac{\partial h(x,\theta)}{\partial \theta} dx \]

CB
 
  • #3
Thank you! It seems the FTOC wouldn't really help me in this case.
 

FAQ: What went wrong when applying the FTOC to maximize arc length?

1. How does increasing the radius of a circle affect the arc length?

As the radius of a circle increases, the arc length also increases. This is because a larger radius means a longer distance around the circle, resulting in a longer arc length.

2. Can the arc length of a circle be greater than the circumference?

No, the arc length of a circle can never be greater than the circumference. The circumference is the total distance around the circle, while the arc length is a portion of that distance.

3. How does the measure of the central angle affect the arc length?

The measure of the central angle is directly proportional to the arc length. This means that as the central angle increases, the arc length also increases, and vice versa.

4. Are there any real-world applications of maximizing arc length?

Maximizing arc length has many practical applications, such as in engineering and construction. For example, it is used to calculate the length of a curved road or the amount of material needed to construct an arch.

5. Can the arc length be negative?

No, the arc length cannot be negative. It is always a positive value representing the distance along the circumference of a circle.

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