Main
2010-03-09 17:34:47
*Reward 1 pin for most got it right before tommor 8am est*
SRL Archive - Mathematics
Calculus with reward
Diddy Kong
2010-03-09 18:37:04
Answer: 4 รท 0 … OH SH*T!
Main
2010-03-09 18:39:42
this is serious man, please help me out here.
Brain
2010-03-09 18:56:11
well lets see…for the first one, take the derivative, and set it to 0, then solve for x
man, this stuff is lame without just having learned it…can't remember much of it.
man, this stuff is lame without just having learned it…can't remember much of it.
XcanadamanX
2010-03-09 18:58:23
for the first one i get 0 as the horizontal tangent
Brain
2010-03-09 18:59:14
0 is one of the horizontal tangents, but just from my glance I would say there is at least one more.
XcanadamanX
2010-03-09 19:08:43
im just going by my crappy calculus software…the one thing canadian i am not proud of :P
Main
2010-03-09 19:13:47
what are you in uwo for? life sci?
XcanadamanX
2010-03-09 19:27:30
who's in uwo? im at mac for eng
Nava2
2010-03-09 21:26:01
I'm at UWO for eng. Currently life sci though. :(
Close though ;)
Close though ;)
HyperSecret
2010-03-10 01:40:43
From: Main_Ftw
*Reward 1 pin for most got it right before tommor 8am est*
I don't have the time to solve them sorry.
A) you will take the derivative of each. d[x^3y^2] + d[4(3x-2y)^3]=d[4y]-d[6x]
d=derivative, you will need to use the product rule on the first, chain rule on the second and the other side is simple.
B) The same thing as A, take the derivative of each operand.
I would have to review for the rest, sorry. I Just don't have that time right now.
ZephyrsFury
2010-03-10 06:35:09
I havent' checked question 2 and I've got the working if you want it (its rather messy though).
Main
2010-03-10 19:01:03
can you show me your steps ? :f:
ZephyrsFury
2010-03-11 05:29:06
Ok..
1. Horizontal tangents when dy/dx = 0,
Find the first derivative function.. simplies to this:
Set dy/dx to 0 and solve for x; only ways to get 0 is if x^2 = 0 or (3 + 4ln(4).x) = 0 as an exponential function can never equal 0.
Hence,
Then you substitute those values into your initial function to get y and since slope = 0, there is no x value in those equations just the y value.
2. Implicitly differentiate them… too hard to explain online tbh =/
3. Again implicitly differentiate and rearrange to get dy/dx = everythingelse. You're giving the value of x = 1, so subsitute that into your y = equation to get the y values, then substitute the sets of coords (you get two y values: (1, 6) and (1, -4)) into your dy/dx function to get your gradients at those points. Then its a simple matter of finding your tangent equations.
4. Key Point: If two lines are perpendicular then their gradients multiply to -1, ie
1. Horizontal tangents when dy/dx = 0,
Find the first derivative function.. simplies to this:
dy/dx = 4^(4x+3).x^2.(3 + 4ln(4).x)
Set dy/dx to 0 and solve for x; only ways to get 0 is if x^2 = 0 or (3 + 4ln(4).x) = 0 as an exponential function can never equal 0.
Hence,
x = 0, x = -3/(4ln(4))
Then you substitute those values into your initial function to get y and since slope = 0, there is no x value in those equations just the y value.
2. Implicitly differentiate them… too hard to explain online tbh =/
3. Again implicitly differentiate and rearrange to get dy/dx = everythingelse. You're giving the value of x = 1, so subsitute that into your y = equation to get the y values, then substitute the sets of coords (you get two y values: (1, 6) and (1, -4)) into your dy/dx function to get your gradients at those points. Then its a simple matter of finding your tangent equations.
4. Key Point: If two lines are perpendicular then their gradients multiply to -1, ie
m1.m2 = -1.
Main
2010-03-11 06:42:50
ty. First one was the one i didn't get, i got 0 but when i graphed it, it was totally different.