1.) values of f(t) are given in the following table:
t 0 2 4 6 8 10
f(t) 137 112 88 68 49 34
Estimate (f prime) f^' (2) and f^' (8)
Please show work so I can understand. I'm really stuck.
if the table comes out messed up, the values are:
(0, 137) (2, 112) (4, 88) (6, 68) (8, 49) (10, 34)
Check what I did for you a minute ago below.
0 137
====== -25
2 112 ****** +1
====== -24
4 88 ******* +4
======= -20
6 68 ******* +1
======= -19
8 49 ********* +4
======== -15
10 34
so
dy/dx is negative
but the change of dy/dx with x is positive right down the table
Between x = 0 and x = 2, y changes -25
dy/dx = -25/2 = -12.5
between x = 2 and x = 4, y changes -24
dy/dx = -24/2 = -12
so averaging to get an instantaneous dy/dx at x = 2
I would say dy/dx = -12.25 at x = 2
go through the same routine at x = 8
Does that help. It is really, really important.
My columns are "change in y" and "change in change of y" for each interval of 2 in x
yeah for 2 but when I tried that for x = 8, I have a huge gap between the two numbers for 8. I got between x=0 and x =8, it is -88. then dividing that by 8, i got -11. then when i did between x= 8 and x =10, i got -55. then dividing that by 8, i got -1.875 and now I'm stuck
it is change in y / change in x
Your change in x is ALWAYS 2 in the table you gave me, not 8
from 6 to 8 dy = -19
so dy/dx = -19/2 = -9.5
from 8 to 10 dy = -15
so dy/dx = -7.5
average at x = 8 is -8.5
so then for x = 8, u do the exact same steps by dividing the numbers by 2?
for x = 8 do u use between x =6 and x=8 and between x=8 and x=10?
Remember dy/dx is the slope at that point, like at x = 8 here
find the slope between x = 6 and x = 8
find the slope between x = 8 and x = 10
average them
sketch what you are doing on a rough graph
calculus. please helppp!! - JJ, Wednesday, February 22, 2012 at 9:08pm
so then for x = 8, u do the exact same steps by dividing the numbers by 2?
calculus. please helppp!! - JJ, Wednesday, February 22, 2012 at 9:11pm
for x = 8 do u use between x =6 and x=8 and between x=8 and x=10?
i tried the for x = 2, -12.25
and for x =8, -8.5 and the website says it's wrong. it doesn't say which one is wrong. should the values be negative or positive?
I think you have it now. I want to make sure because if anything is important in this course, this is it.
its says the answers are wrong. doesn't specify which one. so I'm not sure if the actual answer is wrong or if its the negative sign.
They are both negative.
But your text may not be averaging as I did. They might be just taking one side (a poorer approximation)
so try
-12.5
and
-9.5
we could even get fancier by using the second derivative but I doubt if your text wants you to do that.
the values changed. now its table is: (0, 14) (2, 38)
(4, 60), (6, 77) (8, 93) (10, 104)
so i said the first derivative is positive, second is negative.
f^' (2) ---> between x=0 and x=2, it's 12
between x = 2 and x = 4, it's 11
average = 11.5
f^' (8) ----> between x = 6 and x = 8, it's 8
between x = 8 and x = 10, it's 5.5
average = 6.75
is that right?
agree + then -
agree dy/dx = 11.5 at x = 2
agree 6.75
urghh it says that's not right!
is there a way to use difference quotients for it?
probably can but forget method, hang on
K, using difference quotient you would just use the right side instead of averaging
so (2, 38) to (4,60)
would be
22/2 = 11 so use 11
and
(8,93) to (10,104)
would be
11/2 = 5.5
That is a much rougher approximation than we did but is
[f(x+h)-f(x)]/h