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Derivatives of implicit functions

by courtrigrad
Tags: derivatives, functions, implicit
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courtrigrad
#1
Oct7-05, 03:01 PM
P: 1,237
If you are given the derivative of an implicit function as [tex] y' = \frac{y}{2y+x} [/tex] how would you find all points (x,y) such that the slope at those points is 1/2? Ok so I did: [tex] \frac{y}{2y+x} = \frac{1}{2} [/tex] and got x = 0. So if I substitute x = 0 back into the original equation I get [tex] (0, \sqrt{2}) [/tex]. Would this be the correct method to solve this question? Also show that the slope is never equaled to 0. So I did [tex] \frac{y}{2y+x} = 0 [/tex]. If y = 0, then the original equation wouldnt make sense. Is this a valid response?

Any help is appreciated

Thanks
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Jameson
#2
Oct7-05, 03:39 PM
P: 788
Well, I wouldn't word it "the original equation wouldn't make sense". Can you word this more mathematically or demonstrate why cannot equal 0?
amcavoy
#3
Oct7-05, 03:44 PM
P: 666
Quote Quote by plugpoint
If you are given the derivative of an implicit function as [tex] y' = \frac{y}{2y+x} [/tex] how would you find all points (x,y) such that the slope at those points is 1/2? Ok so I did: [tex] \frac{y}{2y+x} = \frac{1}{2} [/tex] and got x = 0. So if I substitute x = 0 back into the original equation I get [tex] (0, \sqrt{2}) [/tex]. Would this be the correct method to solve this question? Also show that the slope is never equaled to 0. So I did [tex] \frac{y}{2y+x} = 0 [/tex]. If y = 0, then the original equation wouldnt make sense. Is this a valid response?

Any help is appreciated

Thanks
When you solve:

[tex]\frac{y}{2y+x}=\frac{1}{2}\quad (1)[/tex]

and come up with x=0, I don't see how you get the square root of 2. Try plugging x=0 into equation (1). The y variable will cancel out and you get an identity. What does that tell you about the possible values of y?

Alex

Jameson
#4
Oct7-05, 03:46 PM
P: 788
Derivatives of implicit functions

Quote Quote by apmcavoy
I don't see how you get the square root of 2.
He gave y', so I thought he plugged 0 into y, which he didn't give in his post.
mezarashi
#5
Oct7-05, 03:47 PM
HW Helper
P: 661
Quote Quote by apmcavoy
Try plugging x=0 into equation (1). The y variable will cancel out and you get an identity. What does that tell you about the possible values of y?
I think you forgot to note that equation one is the derivative of the function. I think plugpoint may have forgot to post the original equation to where he got the square root 2. The current differential is not separable.
amcavoy
#6
Oct7-05, 03:50 PM
P: 666
Quote Quote by mezarashi
I think you forgot to note that equation one is the derivative of the function. I think plugpoint may have forgot to post the original equation to where he got the square root 2. The current differential is not separable.
You don't need the original equation. If you plug x=0 into y', you get 1/2 no matter what value of y you have. This means that along the vertical line x=0 the slope is the same for all values of y.

Alex


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