Late night Calculus help!

[Poncherelli]

I'm having trouble with a calc problem. Even though i've done quite a few of these kinds of problems, this one is giving me a hard time:

Let f be a function such that f '(x) = (2x+ sin x)/ (x^2 + 1) and f (2)= -1

a. Use Euler's method with 3 equal steps to approx f (2.3).

b. Use f " (x) to show that your approx in part a. is an overstatement of f(2.3).

c. Use the tangent line to the graph of y= f(x) at the point (2, -1) to approx f(2.3).

d. Use the definite integral to approx the value of f(2.3).


Any help is greatly appreciated. Thanks!

 

[Poncherelli]

dam the boards move fast at 12:30 at night... anyone?

 

[Poncherelli]

someone? ...please? ....take a stab at it?

 

[abu]

...... e=mc^2 .....

 

[Poncherelli]

not quite what i'm looking for, but thanks for the effort. Only 6 hours before i leave for school..... so..... keep trying.

 

[goul]

ugh
eulers method, brings back ap calc last year nooo dont want to even think about it
sorry man, if i had my text book id help you, so heres a free bump. hope you get it done

 

[Logix]

Heh... I took AP Calc last year, and I'm in a Calc class again here at college, but I've never heard of Euler's method. My 1100+ page calculus book doesn't have an entry for "Euler" in the index. Sorry, man. But, here's another bump for ya.

 

[silverpig]

I took this class last year. Too bad I remember NOTHING...

a) Which Euler method? I learned of 3. Forward, Back, and Improved... Either way, your prof should have given you a formula for Euler's Method (whichever one...). Just plug 'n' pray

b) My guess here is that you can show that f"(x) will tell you that the function is concave down in this region. Thus, the linear approximation will be overstated (a simple sketch should suffice).

c) The equation of a line is y = mx + b, or y = f'(x)x + b
Find f'(2), and then sub that in for m, put x=2, y=-1 to determine b. Then re use y = mx + b and have x = 2.3, m = same m as above, b = as just determined, y = your approximation value.

d) I'll let someone else take this one. I could figure it out, but I'm kinda lazy and hungry right now.

 

[AaronP]

hahaha I am so glad I finished Calc

 

[Bluga]

this is an initial value problem

consider y' = f(x,y0, y(x0)=y0

then xn = x0 + nh, yn = yn-1 +hf(xn-1, yn-1)

therefore in your case

let

y'= f(x,y)= (2x+ sin x)/ (x^2 + 1) and y(2)= -1
x0 = 2
y0 = -1

we want to estimate f (2.3), setting h = 1/3, use the equation in second line to approximate 3 times.

 

[Poncherelli]

according to Euler's method, i have gotten f(2.3) ~ -0.719 and f"(2.3) -.417

how do i use f"(x) to show my approx is an overestimate from part a?

 

[silverpig]

Read my post above and sketch.

 

[samarth]

use the formula
a.
f (a) ~= f(a) + f'(a) (a-h) + f'' (a) (a-h)^2 / 2 + .... i think this is it from top of my head... it should be in your book
approximating value of f(a) where f(h) is known and 'a' is near 'h'

b.
take second derivative at 'a' to show its concave downwards, therefore approximatio is going 'over' the curve

c
tangent has slope f'(2), use that to approx

d.
dont remember

 

[Poncherelli]

woohoo! c) f(2.3)= -.705

now if anyone can figure out d)! thats the one my buddy told me was difficult and he couldnt get either.

 

[silverpig]

I'm thinking that you'll have to use the fundamental theorem of calculus in there...

 

[Poncherelli]

but i'm only finding one value, so does the fundemental theorem apply? Plus, how do you integrate this thing, i cant substitute and i'm stumped.