# Derivative of velocity then critical points question (Optimization problem)

• November 21st 2009, 10:23 AM
s3a
Derivative of velocity then critical points question (Optimization problem)
Question:
(1 pt) A University of Rochester student decided to depart from Earth after his graduation to find work on Mars. Before building a shuttle, he conducted careful calculations. A model for the velocity of the shuttle, from liftoff at t = 0 s until the solid rocket boosters were jettisoned at t = 63.4 s, is given by
(in feet per second). Using this model, estimate
the absolute maximum value
and absolute minimum value
of the ACCELERATION of the shuttle between liftoff and the jettisoning of the boosters.

My problem: (My work is attached) When I attempt to use the quadratic formula to solve for x on the derivative function of velocity (=acceleration function), the inside of the square root is negative and I therefore cannot take the square root of it. Can someone please help me out?
• November 21st 2009, 12:11 PM
skeeter
Quote:

Originally Posted by s3a
Question:
(1 pt) A University of Rochester student decided to depart from Earth after his graduation to find work on Mars. Before building a shuttle, he conducted careful calculations. A model for the velocity of the shuttle, from liftoff at t = 0 s until the solid rocket boosters were jettisoned at t = 63.4 s, is given by
(in feet per second). Using this model, estimate
the absolute maximum value
and absolute minimum value
of the ACCELERATION of the shuttle between liftoff and the jettisoning of the boosters.

My problem: (My work is attached) When I attempt to use the quadratic formula to solve for x on the derivative function of velocity (=acceleration function), the inside of the square root is negative and I therefore cannot take the square root of it. Can someone please help me out?

I'm using $m$, $n$, $p$, and $q$ to represent those ugly coefficients.

$v(t) = mt^3 + nt^2 + pt + q$

$a(t) = 3mt^2 + 2nt + p$

the graph of $a(t)$ is a parabola. note that $a(t) > 0$ for all $t$, that is why you didn't find any roots. you don't need them anyway.

you want the absolute min and max of $a(t)$ ...

absolute min of $a(t)$ occurs at the parabola's vertex ...

$t = \frac{-2n}{6m}$

absolute max of $a(t)$ ocurs at the parabola's endpoints ... the larger of $a(0)$ or $a(63.4)$
• November 21st 2009, 12:18 PM
s3a
Are you sure you didn't make a mistake because you wrote:

v(t) = mt^3 + nt^2 + pt + q
and they are all + signs when the equation I gave had a - sign before what you called nt^2?
• November 21st 2009, 12:25 PM
skeeter
Quote:

Originally Posted by s3a
Are you sure you didn't make a mistake because you wrote:

v(t) = mt^3 + nt^2 + pt + q
and they are all + signs when the equation I gave had a - sign before what you called nt^2?

no mistake ... letters representing constants need not be positive all the time.

I let $n = -0.08177$ and $q = -12.3$