Finding equation of tangent line

[Dank2]

Homework Statement

The following point (x0,y0), is on the curve sqrtx +sqrty = 1Show that line equation of the tangent line in the point. (x0,y0)
Is x/sqrtx0 + y/sqrty0 = 1

I've found the slope which is
-sqrty/sqrtx.
So slope of the point is -sqrty0/sqrtx0

Homework Equations

The Attempt at a Solution

y-y0=-sqrty0/sqrtx0(x-x0)

Should I get the solution from the equation above?

 

[Christofferk]

Maybe it's easier if you model the problem as one of "one" variable. If you write it as a function y of x, then it is y=x-2\sqrt{x}+1, do you know what the tangent if this is at an arbitrary point? Try to look at the equation for the tangentplane of a \textbf{R}^3-function. The tangent-plane for such an equation at an arbitrary point (x_0,y_0,z_0) is f_{tangent}=\nabla f(x_0,y_0,z_0)\cdot(x-x_0,y-y_0,z-z_0)=0

 

[Dank2]

Haven't det with R3 yet. All we have learned is finding the slope and then using the formula y-y0 = m(x-x0)

 

[Christofferk]

Haven't det with R3 yet. All we have learned is finding the slope and then using the formula y-y0 = m(x-x0)

SOunds as if you know what you're doing, you just need to put the pieces of the puzzle together. :)

 

[Dank2]

SOunds as if you know what you're doing, you just need to put the pieces of the puzzle together. :)

I think need a hint,

 

[Dank2]

Is it like rw

SOunds as if you know what you're doing, you just need to put the pieces of the puzzle together. :)

is it like really easy to show from the equation I've written? I'm not sure which of the 4 variables should I substitute , I can't see how to get even 1 there

 

[Ray Vickson]

Is it like rw

is it like really easy to show from the equation I've written? I'm not sure which of the 4 variables should I substitute , I can't see how to get even 1 there

You do not have 4 variables; you have only 2--namely x and y. These are the two objects that vary along the curve and along the tangent line. The other two objects ($x_0,y_0$) are just some constant input parameters. They do NOT vary along the curve or along the tangent line.

 

[Mark44]

Thread moved. @Dank2, questions involving derivatives should NOT be posted in the Precalc section.

 

[Dank2]

You do not have 4 variables; you have only 2--namely x and y. These are the two objects that vary along the curve and along the tangent line. The other two objects ($x_0,y_0$) are just some constant input parameters. They do NOT vary along the curve or along the tangent line.

I meant I do not know which constant or variable should I subtitude in order to get the right form, from the main equation y= (1-sqrtx)^2 , y0=(1-sqrtx0)^2

 

[Mark44]

Try to look at the equation for the tangentplane of a \textbf{R}^3-function. The tangent-plane for such an equation at an arbitrary point (x_0,y_0,z_0) is f_{tangent}=\nabla f(x_0,y_0,z_0)\cdot(x-x_0,y-y_0,z-z_0)=0

IMO, and this is not a helpful hint.

 

[Mark44]

Show that line equation of the tangent line in the point. (x0,y0)
Is x/sqrtx0 + y/sqrty0 = 1

I've found the slope which is
-sqrty/sqrtx.
So slope of the point is -sqrty0/sqrtx0

The slope at the point $(x_0, y_0)$ is $\frac{-\sqrt{y_0}}{\sqrt{x_0}}$.
You know the point -- $(x_0, y_0)$ -- and you know the slope of the tangent line there -- $\frac{-\sqrt{y_0}}{\sqrt{x_0}}$. Finding the equation of a line through a given point and with a known slope should be straightforward.

 

[Dank2]

The slope at the point $(x_0, y_0)$ is $\frac{-\sqrt{y_0}}{\sqrt{x_0}}$.
You know the point -- $(x_0, y_0)$ -- and you know the slope of the tangent line there -- $\frac{-\sqrt{y_0}}{\sqrt{x_0}}$. Finding the equation of a line through a given point and with a known slope should be straightforward.

Yes it is (y-y_0)=−√y0/√x0*(x-x_0). but i need to show it equals :
x/√x_0 + y/√y_0 = 1

 

[Dank2]

(y-y_0)=−√y0/√x0*(x-x_0) ==> y-y_0 = −√y_0*x/√x0 −√y_0*x_0/√x0
i think that somehow i need to use y=x−2√x+1 from the main equation, i just struggle on this point to do the algebra

 

[Ray Vickson]

Yes it is (y-y_0)=−√y0/√x0*(x-x_0). but i need to show it equals :
x/√x_0 + y/√y_0 = 1

You have $\sqrt{y_0}$ in your first equation and $1/\sqrt{y_0}$ in your second equation. How, then, can you get from the first to the second?

 

[Dank2]

and 1/√y01/y01/\sqrt{y_0} in your second equation.

1/√y_0 ? can't see it

 

[Dank2]

You have $\sqrt{y_0}$ in your first equation and $1/\sqrt{y_0}$ in your second equation. How, then, can you get from the first to the second?

i need a hint

 

[Ray Vickson]

1/√y_0 ? can't see it

You said you wanted to verify that $x/\sqrt{x_0} + y/\sqrt{y_0} = 1$.

 

[Dank2]

You said you wanted to verify that $x/\sqrt{x_0} + y/\sqrt{y_0} = 1$.

that the tangent line equation at the point (x_0, y_0) is equal to it yes.

 

[Dank2]

i need to proof it

 

[Dank2]

You said you wanted to verify that $x/\sqrt{x_0} + y/\sqrt{y_0} = 1$.

i tried coming from the general equation of the tangent line by slope and point, and it didn't come out good.

 

[Dank2]

So it's not trivial?

 

[Mark44]

So it's not trivial?

I think we're waiting for you to finish the problem.

 

[Dank2]

I think we're waiting for you to finish the problem.

I still need a hint regarding the algebra, if that's not asking for much.

 

[Dank2]

I think we're waiting for you to finish the problem.

The issue for me is that from the slope and point equation, I can subtitute y for x, x for y, or y_0 for x_0 or x_0 for y_0 ,
Using the equation y= (1- sqrtx)^2
Or same equation with y_0 and x_0 since this point is on the graph, is that right?

 

[Mark44]

The issue for me is that from the slope and point equation, I can subtitute y for x, x for y, or y_0 for x_0 or x_0 for y_0 ,
Using the equation y= (1- sqrtx)^2
Or same equation with y_0 and x_0 since this point is on the graph, is that right?

Sort of, although what you wrote isn't clear. You can't just replace y with x or x with y, but you can replace y with $(1 - \sqrt{x})^2)$ and you can replace $y_0$ with $(1 - \sqrt{x_0})^2)$.

Your work would be easier to read if you used LaTeX. If you click on any of the expressions I wrote using LaTeX, you can see my script that produces it.

For example, this -- $\sqrt{y_0}$ renders as $\sqrt{y_0}$.

BTW, I'm wondering if there's a mistake in the problem statement. I am getting $\frac{y}{\sqrt{y_0}} + \frac{x}{\sqrt{x_0}} = 2$, not 1 as in post #1. It's possible I have an error.

The problem would be slightly easier if instead of the point $(x_0, y_0)$ we use the point (a, b).

 

[Dank2]

Sort of, although what you wrote isn't clear. You can't just replace y with x or x with y, but you can replace y with $(1 - \sqrt{x})^2)$ and you can replace $y_0$ with $(1 - \sqrt{x_0})^2)$.

Your work would be easier to read if you used LaTeX. If you click on any of the expressions I wrote using LaTeX, you can see my script that produces it.

For example, this -- $\sqrt{y_0}$ renders as $\sqrt{y_0}$.

BTW, I'm wondering if there's a mistake in the problem statement. I am getting $\frac{y}{\sqrt{y_0}} + \frac{x}{\sqrt{x_0}} = 2$, not 1 as in post #1. It's possible I have an error.

The problem would be slightly easier if instead of the point $(x_0, y_0)$ we use the point (a, b).

I will try to use latex once I get to my pc.

Yes I know that, I should have said y in terms of x or y_0 in terms of x_0.

So the right way is trying all possibilities ? Or is there some method I should follow?

And it should equal to 1.

 

[Dank2]

Sort of, although what you wrote isn't clear. You can't just replace y with x or x with y, but you can replace y with $(1 - \sqrt{x})^2)$ and you can replace $y_0$ with $(1 - \sqrt{x_0})^2)$.

Your work would be easier to read if you used LaTeX. If you click on any of the expressions I wrote using LaTeX, you can see my script that produces it.

For example, this -- $\sqrt{y_0}$ renders as $\sqrt{y_0}$.

BTW, I'm wondering if there's a mistake in the problem statement. I am getting $\frac{y}{\sqrt{y_0}} + \frac{x}{\sqrt{x_0}} = 2$, not 1 as in post #1. It's possible I have an error.

The problem would be slightly easier if instead of the point $(x_0, y_0)$ we use the point (a, b).

Ok I've just solved it taking the slope/point equation and the the other , putting both in terms of y, making the nominator equal simplifying . Thanks