# Instantaneous Rate of Change

• Jan 14th 2010, 01:26 PM
agent2421
Instantaneous Rate of Change
Can someone help me solve this problem... I need to calculate the instantaneous rate of change for the following points:

1. 1985
2. 1995
3. 2000
4. 2006

-------------

Based on this data:

Year Public Sector Employment (Number of People in Millions)

1981 2.69009
1982 2.72483
1983 2.74130
1984 2.76072
1985 2.79665
1986 2.84086
1987 2.87816
1988 2.94450
1989 2.96726
1990 3.02734
1991 3.05678
1992 3.06323
1993 3.03748
1994 3.00269
1995 2.95784
1996 2.85133
1997 2.78940
1998 2.77897
1999 2.76987
2000 2.78673
2001 2.81360
2002 2.84346
2003 2.90811
2004 2.94086
2005 2.97973
2006 3.04614

The equation I got for this was y = 0.0002x^3 - 1.3853x^2 + 2763.2x - 2000000

Can you please just help me with the first one... After i understand how to do the first oen I'll know how to do the rest.

Thanks :)
• Jan 14th 2010, 01:54 PM
pickslides
You have

$\displaystyle y = 0.0002x^3 - 1.3853x^2 + 2763.2x - 2000000$

find $\displaystyle \frac{dy}{dx}$ and sub in your points.
• Jan 14th 2010, 01:56 PM
agent2421
Quote:

Originally Posted by pickslides
You have

$\displaystyle y = 0.0002x^3 - 1.3853x^2 + 2763.2x - 2000000$

find $\displaystyle \frac{dy}{dx}$ and sub in your points.

I can't use calculus ... I have to do it algebraically I think... I tried this out but for some reason I get 1 786 499 600 which has to be wrong... I did the calculations twice though and its the same answer...

I really need help for this... I tried it again and I get 1 390 827.59 - 1 296 690.71 for my final answer... but that still is an error for a question like this.... I think i messed up somewhere in the calculations but i'm not sure where.
• Jan 14th 2010, 02:39 PM
pickslides
Quote:

Originally Posted by agent2421
I need to calculate the instantaneous rate of change

This implies the use of calculus
• Jan 14th 2010, 02:50 PM
agent2421
yes but we can also calculate it algebraically... I have not learned calculus yet so I can't use calculus...

We have an equation and then we use the formula:

m = f (x+h) - f (x) / (x+h) - x

= f (x+h) - f (x) / h

I used this formula but I can't get it right...
• Jan 14th 2010, 03:20 PM
pickslides
Quote:

Originally Posted by agent2421

= f (x+h) - f (x) / h

Ok then, show me your workings.
• Jan 14th 2010, 03:25 PM
agent2421
m = 0.002 (1985 + 0.0001)^3 - 1.3853 (1985 + 0.0001)^2 + 2763.2 (1985 + 0.0001) - 2 000 000 - [0.002 (1985)^3 - 1.3853 (1985)^2 + 2673.2 (1985) - 2000000]

= 13699253.6 - 13494541.7

= 204711.6/.0001

= 2 047 117 000

That can't be the answer since I"m looking for instantaneous rate of change... I'm going wrong somewhere but I have no idea where... Can someone please help me for this.
• Jan 14th 2010, 03:27 PM
pickslides
Why did you pick $\displaystyle h = 0.0001$ ?
• Jan 14th 2010, 03:29 PM
agent2421
There's an example of the same type of question in my book and it used h = .0001... they were trying to find the instantaneous rate of change for 2006....

I just followed the same example and used .0001