a particle moves along the y axis so that its position at any time t, for 0 is less than or equal to t which is less than or equal to 5, is given by y(t)=t^4-18t^2. in which intervals is the particle speeding up?

y'=4t^3-36t= 4t(t^2-9)=4t(t+3)(t-3)

y"=12t^2-36=12(t^2-3)=12(t+sqrt3)(t-sqrt3)

so when t=less than sqrt3, y"=+ *- or negagive, slowing down.
t= greater than sqrt3, y"=+ * + speeding up.

Well, it seems our particle is on a wild ride along the y-axis! Let's find out where it's speeding up.

To determine where the particle is speeding up, we need to find the intervals where its velocity is positive and its acceleration is greater than zero.

First, let's find the velocity function by taking the derivative of the position function, y(t).

The derivative of y(t) = t^4 - 18t^2 with respect to t gives us the velocity function v(t) = 4t^3 - 36t.

Now, let's find the acceleration function by taking the derivative of the velocity function, v(t).

The derivative of v(t) = 4t^3 - 36t with respect to t gives us the acceleration function a(t) = 12t^2 - 36.

Since we want to find where the particle is speeding up, we need to find the intervals where a(t) > 0.

To determine the intervals, we need to find the zeros of a(t).

Setting a(t) = 0, we get:

12t^2 - 36 = 0

Dividing both sides by 12, we have:

t^2 - 3 = 0

Adding 3 to both sides, we get:

t^2 = 3

Taking the square root of both sides, we find:

t = ±√3

Now, let's check each interval: 0 ≤ t ≤ 5, for t = 0, t = √3, and t = 5.

For t = 0, a(0) = 12(0)^2 - 36 = -36, so the particle is not speeding up here.

For t = √3, a(√3) = 12(√3)^2 - 36 = 0, so the particle is not speeding up here either.

For t = 5, a(5) = 12(5)^2 - 36 = 288, which is greater than 0.

Therefore, the particle is speeding up on the interval 0 ≤ t ≤ 5.

So hang on tight, because during that time, our particle is accelerating and having a blast along the y-axis!

To determine in which intervals the particle is speeding up, we need to find the intervals where the particle's velocity is positive and its acceleration is also positive.

To find the velocity, we need to take the derivative of the position function y(t) with respect to time:

v(t) = dy(t)/dt = d/dt (t^4 - 18t^2)

Taking the derivative, we get:

v(t) = 4t^3 - 36t

Now, to find the acceleration, we need to take the derivative of the velocity function v(t) with respect to time:

a(t) = dv(t)/dt = d/dt (4t^3 - 36t)

Taking the derivative, we get:

a(t) = 12t^2 - 36

Now we have the velocity and acceleration functions. To find the intervals where the particle is speeding up, we need to find the intervals where both v(t) > 0 and a(t) > 0.

1. Velocity:
To find when v(t) > 0, we set the velocity function greater than zero:

4t^3 - 36t > 0

Factorizing the equation, we get:

4t(t^2 - 9) > 0

This gives us two critical points: t = 0 and t = ±√9.

Plotting these points on a number line, we have:

-∞ |------o------o------o------| ∞
-√9 0 √9

Testing the intervals:
For t < -√9, we choose t = -10:
4(-10)(100 - 9) > 0 => -3600 > 0 => false

For -√9 < t < 0, we choose t = -1:
4(-1)(1 - 9) > 0 => 32 > 0 => true

For 0 < t < √9, we choose t = 1:
4(1)(1 - 9) > 0 => -32 > 0 => false

For √9 < t, we choose t = 10:
4(10)(100 - 9) > 0 => 3600 > 0 => true

Therefore, the intervals where the particle's velocity is positive are (-√9, 0) and (√9, ∞).

2. Acceleration:
To find when a(t) > 0, we set the acceleration function greater than zero:

12t^2 - 36 > 0

Simplifying the equation, we get:

3t^2 - 9 > 0

Factoring out a 3, we have:

3(t^2 - 3) > 0

This gives us one critical point: t = √3.

Plotting this point on a number line, we have:

-∞ |------o------| ∞
√3

Testing the intervals:
For t < √3, we choose t = 0:
3(0^2 - 3) > 0 => -9 > 0 => false

For t > √3, we choose t = 2:
3(2^2 - 3) > 0 => 9 > 0 => true

Therefore, the interval where the particle's acceleration is positive is (√3, ∞).

In conclusion, the particle is speeding up in the interval (√9, ∞).

To determine in which intervals the particle is speeding up, we need to find the intervals where the particle's acceleration is positive.

First, let's find the acceleration of the particle by taking the second derivative of the position function, y(t).

y(t) = t^4 - 18t^2

Taking the first derivative with respect to t gives us the velocity function:

v(t) = 4t^3 - 36t

Taking the second derivative with respect to t gives us the acceleration function:

a(t) = 12t^2 - 36

Now, we need to find the intervals where the acceleration is positive. In other words, we need to find the values of t for which a(t) > 0.

12t^2 - 36 > 0

Dividing both sides of the inequality by 12:

t^2 - 3 > 0

Now, we can solve this quadratic inequality by factoring:

(t - √3)(t + √3) > 0

The critical points are t = -√3 and t = √3. We can set up a number line to test the intervals using test points:

<-----------(-∞)---(-√3)---(√3)---(+∞)------------->

For a test point in the interval (-∞, -√3), let's use t = -4:

(-4 - √3)(-4 + √3) = (-7)(-1) = 7 > 0

For a test point in the interval (-√3, √3), let's use t = 0:

(0 - √3)(0 + √3) = (-1)(1) = -1 < 0

For a test point in the interval (√3, +∞), let's use t = 4:

(4 - √3)(4 + √3) = (7)(1) = 7 > 0

The inequality is true for the intervals (-∞, -√3) and (√3, +∞).

Therefore, the particle is speeding up in the intervals (-∞, -√3) and (√3, +∞).