# Math Help - Application of calculus

1. ## Application of calculus

Could someone please tell me how to calculate the following...?

The angular displacement Θ
radians of a flywheel varies with time $"t"$seconds and follows the equation Θ $= 9t^2 - 2t^2$

Determine:

a)The angular velocity and acceleration of the flywheel when time $"t"$ = 1 Second

b) The time when the angular acceleration is zero.

2. You have the displacement. Surely you know that the velocity is the time derivative of displacement, and that acceleration is the time derivative of velocity...

3. The angular velocity is $\frac{d\theta}{dt}= \frac{d(9t^2- 2t)}{dt}$ and the angular acceleration is the second derivative.

Note: I have assumed that you meant $\theta(t)= 9t^2- 2t$ rather than $\theta(t)= 9t^2- 2t^2$ as you wrote since that would be written as $7t^2$.

The angular displacement Θ radians of a flywheel varies with time $"t"$seconds and follows the equation Θ $= 9t^2 - 2t^3$

Sorry for the Typo

5. The velocity is $\Theta '(t)$ and the acceleration is $\Theta ''(t)$.

6. Sorry I dont follow.

7. Do you know how to differentiate $t^n$ with respect to t?

8. The rate of change of the angular displacement is the angular velocity.

The rate of change of the angular velocity is the angular acceleration
(rate of change of velocity).

Differentiate your displacement expression to get the velocity expression.
Whatever new expression you get for velocity, differentiate that to get the acceleration expression.

9. Right so I have made a start... So far I have worked out that the velocity = $18t -6t2$

Now I have to differentiate this

10. Yes, then you will have 2 expressions for velocity and acceleration.

You will then be able to set t=1 in both of these to answer part (a).

Then onto part (b).

11. Originally Posted by Archie Meade
Yes, then you will have 2 expressions for velocity and acceleration.

You will then be able to set t=1 in both of these to answer part (a).

Then onto part (b).
$\frac{d}{dt} = 18t = 18$ (Does the t stay or go?)

$\frac{d}{dt} = -6t^2 = -12t$

Thus the Velocity = $18 - 12t$

Is this ok so far?

12. Originally Posted by HNCMATHS
$\frac{d}{dt}(18t )= 18$ (Does the t stay or go?)

$\frac{d}{dt}\left(-2t^3\right) = -6t^2 = -12t$

Thus the Velocity = $18 - 12t$

Is this ok so far?
You have a typo.
Your second derivative is the acceleration.

$\frac{d}{dt}(18t)=18\frac{d}{dt}t^1=(18)(1)t^0$

$t^0=1$

so you get

$acceleration=18-12t$

Now write your 2 expressions for velocity and acceleration and try parts (a) and (b) of the question.

13. Originally Posted by Archie Meade
You have a typo.
Your second derivative is the acceleration.

$\frac{d}{dt}(18t)=18\frac{d}{dt}t^1=(18)(1)t^0$

$t^0=1$

so you get

$acceleration=18-12t$

Now write your 2 expressions for velocity and acceleration and try parts (a) and (b) of the question.
Velocity when t = 1 second

$(18)(1)-((6)(1))^2$ Thus Velocity $= -18$

Acceleration when t = 1 second

$(18)-((12)(1)) = 6$

14. Try the velocity again.

You have squared all of $(6t)$

instead of just the "t" part.

15. Originally Posted by Archie Meade
Try the velocity again.

You have squared all of $(6t)$

instead of just the "t" part.
$(18)(1)-((6)(1^2))$ = 12

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