# Math Help - Trying to solve a quadratic to find critical points?

1. ## Trying to solve a quadratic to find critical points?

Hi again guys,

I'm trying to take the derivative of the velocity model of a space shuttle:

v(t) = 0.001895t^3−0.08391t^2+13.69t+2.71

I get the acceleration as:

a(t) = 5.685E-3 t^2 - .16782 t + 13.69

I want to find the critical numbers so I want to set this function equal to zero. I go for the quadratic but I get a negative under the square root. Is there something I'm missing in this problem? Any clarification is appreciated

2. Originally Posted by DannyMath
Hi again guys,

I'm trying to take the derivative of the velocity model of a space shuttle:

v(t) = 0.001895t^3−0.08391t^2+13.69t+2.71

I get the acceleration as:

a(t) = 5.685E-3 t^2 - .16782 t + 13.69

I want to find the critical numbers so I want to set this function equal to zero. I go for the quadratic but I get a negative under the square root. Is there something I'm missing in this problem? Any clarification is appreciated
The graph of $v(t)$ is practically linear.

It does not differ much to $13.69t+2.71$ for $-30

and outside of those $t$ the graph is almost vertical.

It has no turning points.

3. Originally Posted by Archie Meade
The graph of $v(t)$ is practically linear.

It does not differ much to $13.69t+2.71$ for $-30

and outside of those $t$ the graph is almost vertical.

It has no turning points.
The question sets the interval between t= 0 and t= 38.9 and asks for the abs max and min acceleration.

If there are no critical numbers should I just use the closed interval method from the texbook?

4. Originally Posted by DannyMath
The question sets the interval between t= 0 and t= 38.9 and asks for the abs max and min acceleration.

If there are no critical numbers should I just use the closed interval method from the texbook?
The graph of acceleration has a turning point as it is quadratic.

The derivative is positive across that interval of $t$.

The derivative is U-shaped so, the acceleration is a minimum, where the 2nd derivative is 0.

The maximum value of acceleration occurs at $t=38.9$