How to approximate instantaneous rate of change

[Raerin]

The function for attitude vs air pressure is f(x) = 2100/x

1. Approximate the instantaneous rate of change at 100 millibars
a) Use the equation to calculate the point (100,___)

I found the y-value to be 21.

b) We need two points to calculate a slope, so to approximate the tangent line we use the equation to calculate a point very close to 100: (100.00001, ______) or (99.9999,_____)


So for the y-value are you supposed to find a number close to 21? But how do you decide what number is the most appropriate? Do you just put in any number of decimals you want?

I did 21.00001 where the slope equals 1. But, apparently slope is supposed to be -0.21. I have no idea how you come to that conclusion.

 

[Jameson]

To approximate the slope there you can use:

\(\displaystyle \frac{f(x_2)-f(x_1)}{x_2-x_1}\)

where $x_2$ is very slightly larger than the true $x$ you are looking at and $x_1$ is slightly smaller. What do you get now?

 

[Raerin]

To approximate the slope there you can use:

\(\displaystyle \frac{f(x_2)-f(x_1)}{x_2-x_1}\)

where $x_2$ is very slightly larger than the true $x$ you are looking at and $x_1$ is slightly smaller. What do you get now?

I know that, but how do you determine the number that is slightly larger? I did:

21.00001 - 21/100.00001 - 100

This gave me a slope of 1, but the slope is supposed to be -0.21. I don't know how you get the correct answer.

 

[Jameson]

I know that, but how do you determine the number that is slightly larger? I did:

21.00001 - 21/100.00001 - 100

This gave me a slope of 1, but the slope is supposed to be -0.21. I don't know how you get the correct answer.

They gave these to you. 100.00001 and 99.9999. Plug those into $f(x)$ to find the numerator. What are $f(100.00001)$ and $f(99.9999)$?

 

[Raerin]

For f(100.00001) I got 20.9999979
For f(99.9999) I got 21.000021

Okay, I got the answer.

So, what if they didn't give me a number? what do I do then?

 

[Jameson]

Well if this is for a calculus class then you won't have to use approximations much longer as you'll be able to calculate the true instantaneous slope but if you aren't given any test values for $x$ then I would start with adding and subtracting .0001 or so and then see how that answer looks. If it doesn't seem to be close to a certain value then maybe try adding and subtracting something smaller.