Calculus help

[WhiteXFire]

I haven't used this crap in like 5 years, but I'm still waiting for the hard ones...

Actually, now that I think back, I had Calc III either freshman or sophomore year, went to class only on exam days, and still got an A, haha. It's all about reading/remembering the rules from the book.

[boostedfury]

did u have Murda Jome?

[J.Ralli]

yes i did lmao

[boostedfury]

ok ok lol touche: Find the equation of the tangent line to the graph of the function y^2 + ln(xy) = 2 at the point (e,1)

[boostedfury]

haha see I am not getting a curve like that dude gave, I am straight up earning my failing grades hahahah

[CleanNeon98]

Originally Posted by omgjacki

We had rep points for like a minute. If we still had them Cleanneon would be banned already =P JK
But I would give WhiteXFire lots of points, too.

thats not nice

Actually I liked the system, and not just cause I was getting good rep

[J.Ralli]

Originally Posted by boostedfury

haha see I am not getting a curve like that dude gave, I am straight up earning my failing grades hahahah

see the curve made it better, the only way to get an F was by cursing him out, blank paper got you a D, get atleast 1 right and you get a C... you get the idea lol

[boostedfury]

thats BULL i got a 12 week course squished into 6 weeks of summer classes, and NO CURVE lol...yea I am pissed at J. ralli haahhah

[omgjacki]

Originally Posted by J.Ralli

see the curve made it better, the only way to get an F was by cursing him out, blank paper got you a D, get atleast 1 right and you get a C... you get the idea lol

I like that system. Ha

[J.Ralli]

this kid pissed me off once tho... got like 40% right n threw off the curve, he got an A while the rest of us got a D

[boostedfury]

who wouldnt right hahah..o does anyone have the daily times for Monday...apparently I made the paper for deans list at PSU..let me know i would like to read it..

[WhiteXFire]

Where's a TI-83 when you need one...

[boostedfury]

haha i wish i had batteries for mine

[WhiteXFire]

Originally Posted by boostedfury

ok ok lol touche: Find the equation of the tangent line to the graph of the function y^2 + ln(xy) = 2 at the point (e,1)

I know how to do this, I'm just trying to remember what i need to do to find the derivative of that function. A tangent line to a curve, provided there are no breaks or sharp edges at the point given, is defined by the point (x, f(x)) and the slope given as f'(x), which in this case is (e,1) and f'(e). The equation for the line is y - y0 = m(x - x0) where m is the slope and (x0,y0) is a point on the line, so you'll have y = f'(e)(x - e) + 1.

[boostedfury]

awesome awesome awesome...got one more for you whitefire that I know will be soo easy..its just sketching a graph

[boostedfury]

make a rough sketch of the function with these characteristics: f(0)- f(2)= 0, f prime (x)<0 for x<1, f prime (1)=0, f prime(x)> 0for x>1, f dbl prime (x)>0 for all values of x

[[sr20]sean]

Originally Posted by WhiteXFire

Where's a TI-83 when you need one...

ti84 work? lol

i havnt taken calc yet just pre-calc

now i know what im going to be looking at for calc

[WhiteXFire]

Originally Posted by boostedfury

ok ok lol touche: Find the equation of the tangent line to the graph of the function y^2 + ln(xy) = 2 at the point (e,1)

Originally Posted by WhiteXFire

I know how to do this, I'm just trying to remember what i need to do to find the derivative of that function. A tangent line to a curve, provided there are no breaks or sharp edges at the point given, is defined by the point (x, f(x)) and the slope given as f'(x), which in this case is (e,1) and f'(e). The equation for the line is y - y0 = m(x - x0) where m is the slope and (x0,y0) is a point on the line, so you'll have y = f'(e)(x - e) + 1.

Ok, so you need to find the derivative with respect to x of y^2 + ln(xy) = 2.
d[y^2 + ln(xy)]/dx = d[2]/dx
d[y^2 + ln(x) + ln(y)]/dx = 0
2yy' + 1/x + y'/y = 0
y'(2y + 1/y) = -1/x
y' = -1/(x(2y + 1/y))
y' = -1/(2xy + x/y)
y'(e) = -1/(2ey + e/y)
Don't forget that y = f(x) = f(e) = 1.
y'(e) = -1/(2e + e)
y'(e) = -1/(3e)

Putting this back into the equation, you get:
y = (-1/(3e))(x - e) + 1
y = (e-x)/(3e) + 1
y = 1/3 - x/(3e) + 1

y = -x/(3e) + 4/3

At least I think, lol.

[boostedfury]

thanks for all teh help whitefire that was awesome of you. now i am gonna get some sleep and I will let you know how I do on it

[TGilb2007]

WTF??? Ok, thankfully, in the 4 years I went to college, I NEVER took Calculus.... that isnt even real math, is it?


And for the record, I am kidding about the real math part.