Some1 Plz Help Me With This Impossible Calc Problem
find the slope of the tangent line to the curve
cos(5x-4y)-xe^(-x)=-5pi/2 at the point (5pi/2, 2pi)
i did this like 10 times, each getting a different wrong answer
can any of u guys try it and tell me what u got
October 25, 2012
11 Comments • Newest first
roxcibop and allhailmoi2, we all had like the same way to do it, but our answers are wrong. does that mean we're supposed to do it a nother way?
ROXCIBOP IS A GENIUS!
[quote=achyif]When you put it into wolframalpha are you sure you typed it correctly?[/quote]
yes cuz in wolfram it shows you what u put and its the same
btw this is wat wolfram gives me
e^(-x) (-1+x-5 e^x sin(5 x-4 y))
When you put it into wolframalpha are you sure you typed it correctly?
[quote=allhailmoi2]cos(5x-4y)-xe^(-x)=-5pi/2 -->implicit differentiation
[-sin(5x-4y)](5-4dy/dx) - [e^-x + (-1)(xe^-x)] = 0
[-sin(5x-4y)](5-4dy/dx) -e^-x +xe^-x = 0
[-sin(5x-4y)](5-4dy/dx) = e^-x +xe^-x
(5-4dy/dx) = (e^-x +xe^-x) / [-sin(5x-4y)]
-4dy/dx = { (e^-x +xe^-x) / [-sin(5x-4y)] } -5
[b]dy/dx = { (e^-x +xe^-x) / [4sin(5x-4y)] } +5/4[/b]
and then sub in dy/dx = {[e^-5pi/2](1+x) / 4} +5/4
which is like 1.2508
LOOL this is probably wayyy off...but if your teacher says yur in the right direction, it means yur in the right direction...soo keep at it
Soooo.....this is my one an only attempt at such a stupid and outrageously tedious question like that...[/quote]
in my last attempt my dy/dx is similar to urs but i get subtraction for the numerator, but i think i got ur equation b4. w/e let me try it
...ur answre was wrong too failed 12 times alraedy
well me ap calc teacher said that the way i was doing it was right but i got it wrong o.o
i tried this problem 9 times already
try google ask a teacher it helps 100% but only up to 9:00pm its a online chat thing
this question is worthy enough for god to anwer
pray to god for the answer.
i tried wolframalpha and i got it wrong
idk but whenever i got stuck www.wolframalpha.com like a baws
The tangent slope to the curve? Are sure the wording isn't [b]find the line tangent to the curve?[/b]