Slope of Curve at Point (p,q): Solving Problem

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Homework Help Overview

The discussion revolves around finding the slope of a curve defined by the equation √x + √y = 1 at a specific point (p, q). Participants are exploring the process of rearranging the equation, taking derivatives, and applying tangent line equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss rearranging the curve equation to isolate y and taking its derivative. There are attempts to apply implicit differentiation and substitute specific values for p and q. Some participants express confusion about the correctness of their derivatives and tangent equations.

Discussion Status

The discussion is ongoing, with various approaches being explored, including implicit differentiation and checking the rearrangement of the original equation. Participants are questioning their methods and the results they are obtaining, indicating a lack of consensus on the correct path forward.

Contextual Notes

There is a reference to the relationship √p + √q = 1, which is a constraint that may influence the calculations. Participants are also struggling with the implications of their derived equations and how they relate to the expected answers.

SteliosVas
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Here is a good question, been trying to work it out all evening, and were all engineering students, and struggling!

Points (p,q) lie on the curve √x + √y = 1

Rearranging to make y the subject we get y=(√x +1)2

We than take the derivative which gives us 1-[itex]\frac{1}{√x}[/itex]

Than since we know x=p and y=q

Plugging in p into y' we get 1-[itex]\frac{1}{√p}[/itex]

Than into tangent equation y-y1=m(x-x1)

Giving us y - q = 1-[itex]\frac{1}{√p}[/itex] (x-p)

Now compared to the answer which says its √pq or x√q + y√p

I am so lost?

I have attached a picture as well to help! :confused:
 

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For starters, check that you have rearranged & taken the derivative correctly. Maybe doing dx/dy, as well, will get you somewhere. Just guessing!
 
SteliosVas said:
Here is a good question, been trying to work it out all evening, and were all engineering students, and struggling!

Points (p,q) lie on the curve √x + √y = 1

Rearranging to make y the subject we get y=(√x +1)2

We than take the derivative which gives us 1-[itex]\frac{1}{√x}[/itex]

You could try Implicit Differentiation.
 
Something else that might help is:

[tex]p\sqrt{q} + q\sqrt{p} = \sqrt{pq}(\sqrt{p}+\sqrt{q})[/tex]
 
PeroK said:
You could try Implicit Differentiation.

So I did implict differention of [itex]\sqrt{x}[/itex] + [itex]\sqrt{y}[/itex]= 1

and get [itex]\frac{dy}{dx}[/itex]=[itex]\frac{-\sqrt{y}}{\sqrt{x}}[/itex]

Knowing x=p and y=q

I sub in p (as x) to get m (the gradient of the tangent)

When substituting this into the tangent equation, I still don't get a similar answer.

Am I going wrong somewhere?
 
Last edited:
Remember that ##\sqrt{p} + \sqrt{q} = 1##.
 
vela said:
Remember that ##\sqrt{p} + \sqrt{q} = 1##.

Okay so I have no idea anymore.

I have got the implicit derivative as [itex]\frac{-\sqrt{y}}{x}[/itex].

Therefore the gradient of the tangent to the curve ([itex]\sqrt{x}[/itex]+[itex]\sqrt{y}[/itex]=1 at points p,q is:

Now as I said earlier, x1=p and y1=q

m= [itex]\frac{-\sqrt{p}}{\sqrt{q}}[/itex]

Substituting all known information into the tangent equation of (y-y1)=m(x-x1)

We get y - q = [itex]\frac{-\sqrt{p}}{\sqrt{q}}[/itex] * (x - p)

From here, is where I got no idea?
 
ImageUploadedByPhysics Forums1409547175.334908.jpg


I think this is correct?
 
Last edited:

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