Is the Magnitude of the Fly's Acceleration Constant in a Helical Path?

[Ravenatic20]

I hope I posted in the right place. Sorry in advanced.

Homework Statement

A buzzing fly moves in a helical path given by the equation
r(t) = ib sin \omegat + jb cos \omegat + kct^{2}
Show that the magnitude of the acceleration of the fly is constant, provided b, \omega, and c are constant.

The Attempt at a Solution

x = b sin \omegat
y = b cos \omegat
z = ct^{2}

In class we did a similar problem, but in that problem we had to find the trajectory in space. I'm a little slow, but it's just not helping me with this one. Same with the textbook. I'll go to my teacher if I have to.

I'm not asking someone to do the problem, just get me started. Once that happens, I'll try to go over it here in case I have more questions. Thank you!

 

[Dick]

The acceleration is the second derivative of r(t) with respect to t. What is that?

 

[Ravenatic20]

The acceleration is the second derivative of r(t) with respect to t. What is that?

The first derivative:
ib\omega cos \omegat - jb\omega sin \omegat + 2kct

Second derivative:
-ib\omega^{2} sin \omegat - jb\omega^{2} cos \omegat + 2kc

Is that right?

 

[Nabeshin]

Now find the magnitude...

 

[gabbagabbahey]

Second derivative:
-ib\omega^{2} sin \omegat - jb\omega^{2} cos \omegat + 2kc

Is that right?

Sure; that;s correct ...But very ugly! Try writing the entire equation inside the [ tex] or [ itex] tags instead:

\mathbf{a}(t)=-b\omega^2\sin(\omega t)\mathbf{i}-b\omega^2\cos(\omega t)\mathbf{j}+2c\mathbf{k}

(You can click on the above equation to see the code that generated it)

Now, as Nabeshin said, calculate the magnitude

 

[Ravenatic20]

Thank you guys :)

Now what is the first step in calculating the magnitude? I am used to plugging in numbers to do that.

 

[gabbagabbahey]

You know what the x,y, and z-components of a are, so square them, add the squares, and take the square root as per usual.

||\mathbf{a}||=\sqrt{a_x^2+a_y^2+a_y^2}

 

[Ravenatic20]

So I take \mathbf{a} (the second derivative above), and factor in x for the first part. So it would look like this:

\mathbf{a_x}(t)=[-b\omega^2\sin(\omega t)\mathbf{i}-b\omega^2\cos(\omega t)\mathbf{j}+2c\mathbf{k}*\mathbf{b}sin(\omega t)]^2, where \mathbf{b}sin(\omega t) is x.

Then do the same thing for y and z, add the terms up, and take the square root. Am I on the right track? Then I simplify as much as possible?

 

[gabbagabbahey]

So I take \mathbf{a} (the second derivative above), and factor in x for the first part. So it would look like this:

\mathbf{a_x}(t)=[-b\omega^2\sin(\omega t)\mathbf{i}-b\omega^2\cos(\omega t)\mathbf{j}+2c\mathbf{k}*\mathbf{b}sin(\omega t)]^2, where \mathbf{b}sin(\omega t) is x.

Then do the same thing for y and z, add the terms up, and take the square root. Am I on the right track? Then I simplify as much as possible?

Huh?!

No! a_x is the x-component of a...that's just -b\omega^2\sin(\omega t)...what are
a_y and a_z?

 

[Ravenatic20]

Huh?!

No! a_x is the x-component of a...that's just -b\omega^2\sin(\omega t)...what are
a_y and a_z?

Thank you. I always over complicate things. I believe I know how to do it now.

To answer your question, y is -b\omega^2\cos(\omega t) and z is 2c
Now I square them, add them up, and take the square root of that sum (this:||\mathbf{a}||=\sqrt{a_x^2+a_y^2+a_y^2}). Correct?

 

[D H]

Correct. So what is the result?