Homework Help: Finding equation of tangent line

1. Dec 10, 2016

Dank2

1. The problem statement, all variables and given/known data
The following point (x0,y0), is on the curve sqrtx +sqrty = 1

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

2. Relevant equations

3. The attempt at a solution
y-y0=-sqrty0/sqrtx0(x-x0)

Should I get the solution from the equation above?

2. Dec 10, 2016

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 $f_{tangent}=\nabla f(x_0,y_0,z_0)\cdot(x-x_0,y-y_0,z-z_0)=0$

3. Dec 10, 2016

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)

4. Dec 10, 2016

Christofferk

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

5. Dec 10, 2016

Dank2

I think need a hint,

6. Dec 10, 2016

Dank2

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

7. Dec 10, 2016

Ray Vickson

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.

8. Dec 10, 2016

Staff: Mentor

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

9. Dec 10, 2016

Dank2

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

Last edited by a moderator: Dec 10, 2016
10. Dec 10, 2016

Staff: Mentor

IMO, and this is not a helpful hint.

11. Dec 10, 2016

Staff: Mentor

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.

12. Dec 11, 2016

Dank2

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

Last edited: Dec 11, 2016
13. Dec 11, 2016

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

14. Dec 11, 2016

Ray Vickson

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?

15. Dec 11, 2016

Dank2

1/√y_0 ? can't see it

Last edited: Dec 11, 2016
16. Dec 11, 2016

Dank2

i need a hint

17. Dec 11, 2016

Ray Vickson

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

18. Dec 11, 2016

Dank2

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

19. Dec 11, 2016

Dank2

i need to proof it

20. Dec 11, 2016

Dank2

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

21. Dec 11, 2016

Dank2

So it's not trivial?

22. Dec 11, 2016

Staff: Mentor

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

23. Dec 11, 2016

Dank2

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

24. Dec 11, 2016

Dank2

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?

25. Dec 11, 2016

Staff: Mentor

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).