Finding equation of tangent line

  • Thread starter Dank2
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  • #1
Dank2
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Homework Statement


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



Homework Equations




The Attempt at a Solution


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

Should I get the solution from the equation above?
 

Answers and Replies

  • #2
Christofferk
18
1
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 [itex]y=x-2\sqrt{x}+1[/itex], do you know what the tangent if this is at an arbitrary point? Try to look at the equation for the tangentplane of a [itex]\textbf{R}^3[/itex]-function. The tangent-plane for such an equation at an arbitrary point [itex](x_0,y_0,z_0)[/itex] is [itex]f_{tangent}=\nabla f(x_0,y_0,z_0)\cdot(x-x_0,y-y_0,z-z_0)=0[/itex]
 
  • #3
Dank2
213
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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
Christofferk
18
1
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. :)
 
  • #5
Dank2
213
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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,
 
  • #6
Dank2
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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
 
  • #7
Ray Vickson
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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.
 
  • #8
36,342
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Thread moved. @Dank2, questions involving derivatives should NOT be posted in the Precalc section.
 
  • #9
Dank2
213
4
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
 
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  • #10
36,342
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Try to look at the equation for the tangentplane of a [itex]\textbf{R}^3[/itex]-function. The tangent-plane for such an equation at an arbitrary point [itex](x_0,y_0,z_0)[/itex] is [itex]f_{tangent}=\nabla f(x_0,y_0,z_0)\cdot(x-x_0,y-y_0,z-z_0)=0[/itex]
IMO, and this is not a helpful hint.
 
  • #11
36,342
8,297
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.
 
  • #12
Dank2
213
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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
 
Last edited:
  • #13
Dank2
213
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(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
Ray Vickson
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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?
 
  • #15
Dank2
213
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and 1/√y01/y01/\sqrt{y_0} in your second equation.

1/√y_0 ? can't see it
 
Last edited:
  • #16
Dank2
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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
 
  • #17
Ray Vickson
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1/√y_0 ? can't see it

You said you wanted to verify that ##x/\sqrt{x_0} + y/\sqrt{y_0} = 1##.
 
  • #18
Dank2
213
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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.
 
  • #19
Dank2
213
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i need to proof it
 
  • #20
Dank2
213
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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.
 
  • #21
Dank2
213
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So it's not trivial?
 
  • #23
Dank2
213
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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.
 
  • #24
Dank2
213
4
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?
 
  • #25
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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).
 
  • #26
Dank2
213
4
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.
 
  • #27
Dank2
213
4
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
 

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