Finding the point on the graph where the tangent is parallel

[meeklobraca]

Homework Statement

Find the point on the graph y = sqrt 2x-1 where the tangent line is parallel to x - 3y = 16

Homework Equations

The Attempt at a Solution

So what I started off was was trying to find the slope of x - 3y = 16. I rearranged the equation to a y = equation and got 16-x /-3 which I got a slope of -3.

Then I went to find the deriviative of y = sqrt 2x-1 which I get (x - 1/2)exp-1/2

From there i get a little lost. The derivative = the slope of the tangent which would get me my x and y points, but my x values get a lot wonky. Would somebody be able to lead me in the right direction of finding my x point?

Thanks!

 

[gabbagabbahey]

So what I started off was was trying to find the slope of x - 3y = 16. I rearranged the equation to a y = equation and got 16-x /-3 which I got a slope of -3.
!

Be careful with your negative signs!

[tex]y=16-\frac{x}{-3}=16+\frac{x}{3}[/tex]

so your slope is ___?

Then I went to find the deriviative of y = sqrt 2x-1 which I get (x - 1/2)exp-1/2

You should double check your derivative too!

 

[meeklobraca]

OKay, with that then I get a slope of 1/3, I am still not clear on the derivative though? If its
(2x-1)exp 1/2, wouldn't that be (1/2)(2x-1)exp-1/2?

 

[gabbagabbahey]

OKay, with that then I get a slope of 1/3, I am still not clear on the derivative though? If its
(2x-1)exp 1/2, wouldn't that be (1/2)(2x-1)exp-1/2?

You need to use the chain rule:

[tex]\frac{d}{dx} (2x-1)^{1/2}=\frac{1}{2}(2x-1)^{-1/2}\cdot\frac{d}{dx}(2x-1)[/tex]

 

[meeklobraca]

Okay, this is what I've got from your equation.

1/2(2x-1)-1/2 x 2

(2x-1)-1/2

So with that then I have the deriviative equaling the slope of the tangent which I've figured to be 1/3.

How do I get rid of that negative exponent?

 

[meeklobraca]

To answer my own question, I took the reciprocal of the 2x-1 line so I got

1/(2x-1) = 1/9 A little cross multiplication and I landed x = 5?

Throw that into the equation and we get a y value of 3, therefore the answer is (5,3)?


haha, sorry bud I am editting this as your trying to reply, it dawned on me in like 5 seconds, what did I do with the square root??

 

[gabbagabbahey]

To answer my own question, I took the reciprocal of the 2x-1 line so I got

1/(2x-1) = 1/9 A little cross multiplication and I landed x = 5?

Throw that into the equation and we get a y value of 3, therefore the answer is (5,3)?

Looks good to me

 

[meeklobraca]

Be careful with your negative signs!

[tex]y=16-\frac{x}{-3}=16+\frac{x}{3}[/tex]

How come you got 16 + x/3 and not 16-x/-3?

Wouldnt diving 16-x by -3 equal 16/-3 - x/-3?

 

[HallsofIvy]

He was assuming you meant what you said!

16- x/-3 means 16- (x/-3). If you mean (16- x)/-3 you should have used parentheses to make that clear.

 

[meeklobraca]

So which is it that I go from?

If you have x - 3y = 16 how do you adjust that to make it a y = equation?

 

[Dick]

x-3y=16 -> x-16=3y -> (x-16)/3=y -> x/3-16/3=y. This shouldn't be the hardest part of the problem. gabba didn't check the derivation from the start.

 

[meeklobraca]

If you wouldn't mind confirming for me, but I don't think that changes anything, my answer of (5,10) is still correct?

 

[Dick]

If you wouldn't mind confirming for me, but I don't think that changes anything, my answer of (5,10) is still correct?

x=5 is good. But if y=sqrt(2x-1) what does that make y?

 

[meeklobraca]

Yup, sorry I misread my answer, my answer was (5,3)

 

[Dick]

Yup, sorry I misread my answer, my answer was (5,3)

That's fine.