Hard Calculus Word Problem

Tony Zalles

Hi,

Ok I'm havin troubles trying to solve this calculus problem.

-Rolling a Ball Bearing-
A ball bearing is placed on an incline plane and begins to roll. The angle of elevation of the plane is &theta;.
The distance (in meters) the ball bearing rolls in t seconds is

s(t) = 4.9(sin &theta;)t^2

a. Determine the speed of the ball bearing.

b. What value of &theta; will produce the maximum speed at a particular time.

Ok so here is how I "tackled" the problem.

ok part a, asks for speed so I take the derivative of the function to get velocity.

v(t) = (0)(sin &theta;)(t^2) + (4.9)(cos &theta;)(t^2) + (4.9)(sin &theta;)(2t)
v(t) = 4.9(t^2)(cos &theta;) + (9.8t)(sin &theta;)

ok now from here I'm not sure if my next steps are right...

Ok since they asked for the speed, i'm assuming they want an actual value....versus what I have thus far....

ok now setting what I have for v(t) to zero, I solve for t,

thus I get: t = -2tan &theta;

Now I take that value and put it back in to my original equation.

Now I have s(t) = 4.9(sin &theta;)(-2tan &theta;)^2

Now I solve for &theta;, by setting s(t) equal to zero.

but I end up with this....

4 = (cos &theta;)^2, and I don't think this is the right answer...

So any help on this would be nice,

Thanks.

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Hurkyl

Staff Emeritus
Gold Member
ok part a, asks for speed so I take the derivative of the function to get velocity.
In particular, you're taking the derivative with respect to t.

&theta; is a constant with respect to t... try to differentiate again.

Ok since they asked for the speed, i'm assuming they want an actual value....versus what I have thus far....
Nope. Aside from your mistake, they want what you had.

ok now setting what I have for v(t) to zero, I solve for t,
Why set v(t) to 0? You're trying to maximize v(t) with respect to &theta;, right? You want to set zero equal to the derivative of v(t) with respect to &theta;.

Tony Zalles

In particular, you're taking the derivative with respect to t.

&theta; is a constant with respect to t... try to differentiate again..
hmn...yea actually right after I posted, I wondered if I was differentiating wrong.

Ok so the correct differentiation should be as follows.

v(t) = 9.8(cos &theta;)(sin &theta;)t

therefore thats part A right there.

Part B than should be the derivative of part A, set equal to zero.

Code:
a(t) = (sin &theta;)(cos &theta;)
0 = (sin &theta;)(cos &theta;)
&theta; = 45 Degrees
Ok, I hope i got it right this time.

Thanks,

Hurkyl

Staff Emeritus
Gold Member
Your derivative is still wrong... try writing out all of the steps (like you did when you first learned derivatives) and maybe you'll see your mistake... or at least I'll be able to see what you did wrong when you post it.

HallsofIvy

Homework Helper
For the first part of this problem &theta; is CONSTANT. What is the derivative of at2? Now replace a with 4.9cos(&theta;).

After you have that velocity (at any time t), you want to start thinking "what would happen if we changed &theta;?". Specifically, what value of &theta; gives the maximum velocity. Now t is fixed and &theta; is the variable!

rhit2013

Well, your first mistake is using the velocity function as your speed function. Velocity IS NOT the same as speed. Speed is the magnitude of velocity. You should probably fix that first. Everything else will fall into place as you go along.

Redbelly98

Staff Emeritus
Homework Helper
Welcome to Physics Forums Hopefully Tony Zalles is not still working on this problem after more than 6 years. However, note that the ball simply moves down the slope (starting from rest) without reversing direction. So the speed-velocity distinction is not important here.

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