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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: sally baker on September 27, 2011, 01:34:22 pm

Title: unit 2 methods help
Post by: sally baker on September 27, 2011, 01:34:22 pm

How do you find the turning point of this equation.

v(t)= 5/8(10t^2 - t^3/3)   

Title: Re: unit 2 methods help
Post by: b^3 on September 27, 2011, 01:37:58 pm

Find when the derivative equals 0.
i.e. f'(x)=5/8(20t-t²)=0
20t-t²=0
t(20-t)=0
so t=0 or t=20
So it has two turning points. No find he y-coordinate.
f(0)=0, so point (0,0)
f(20)=833+1/3
(20,833+1/3)

Title: Re: unit 2 methods help
Post by: sally baker on September 27, 2011, 01:38:38 pm

can someone help me with this question.
 

A particle moves along a straight line so that after t seconds its distance from O , a fixed point on the line , is s m , where s=t^3-3t^2+2t


A) where is the particle at  O?
b) what is its velocity and acceleration at these times?
c)what is the average velocity  during the first second?

Title: Re: unit 2 methods help
Post by: sally baker on September 27, 2011, 01:39:57 pm

no i did that , but to draw the graph what will its maximum and minimum be? 

Title: Re: unit 2 methods help
Post by: b^3 on September 27, 2011, 01:43:17 pm

Plug in a value of either side of the turning point into the derivative to find out the gradient. If it is positive on the left and negative on the right it is a maximum and vice versa. That makes the (0,0) a minimum and the other one a maximum.

Title: Re: unit 2 methods help
Post by: sally baker on September 27, 2011, 01:45:43 pm

arent the 0 and 20 the x values but it says on the graph ,  0 and 20 are x vvalues and the turning point is (10,62.5 ) thanks anyways