Question

On May 7, 1992, the space shuttle Endeavour was launched on mission to install a new motor
in an Intelsat communications satellite. The equation describing its velocity on the time
interval to [0,125] is given by
v(x) = .001461x^3-.11553x^2+24.9817x-21.2687
Use the equation to find a model for acceleration. Estimate the maximum and minimum values of the
acceleration during the first 125 seconds

Answer 1

x is the variable for distance?

Answer 2

im pretty sure x is time and my calc teacher just screwed up by changing t's to x on accident

Answer 3

if its time,then it is a pretty simple question.do you know what does acceleration mean.the answer lies in the definition of acceleration

Answer 4

acceleration is just the change of speed over the change in time.

Answer 5

correct.so now here you r given the velocity and the question is asking you to find the rate of change of velocity with respect to time(acceleration).to find it,you only have to differentiate v(x) with respect to x

Answer 6

So just take the derivative of the equation then?

Answer 7

yes

Answer 8

that will give you the expression for acceleration

Answer 9

for finding the maxima and minima you can use the second derivative test

Answer 10

I was sick the last day of class so I don't know what the second derivative test is

Answer 11

when you have an expression,and you gotta find its extrema,the first step is to find its first derivative and set it equal to zero.find the value of x when f'(x)=0.this gives you the critical points.now these critical points are those where the maxima or minima occurs.so they should lie in the interval given in the question.the next step is to find the second derivative of f(x) and find f''(c) where c is a critical point and use the following points http://www.mathwords.com/s/second_derivative_test.htm

Answer 12

Alright I knew about the extrema just didnt know about taking the second derivative.

Answer 13

http://www.math.hmc.edu/calculus/tutorials/secondderiv/ this one is good

Answer 14

still any doubt?

Answer 15

taking the second derivative is easy. Its the second test thats throwing me

Answer 16

did you check out the link i provided above.it explains the test graphically and makes it easier to understand

Answer 17

I understand I'm taking the critical points of the first test and applying them to the second right?

Answer 18

exactly

Answer 19

So now I just have to factor and find the critical points for the first derivative then use them for the 2nd derivative

Answer 20

yes

Answer 21

do i use the points 0 and 125 in the second derivative as well?

Answer 22

when you've found out the critical point,you've to check whether these points lie in the interval [0,125].because only when these points lie in the given interval,can you find the maxima and minima at these points

Answer 23

I found the critical point at 26.35865845 just plugging that 0 and 125 in now and will find max and min from there

Answer 24

yes

Answer 25

Alright I found the answer thank you very much for all of your help.