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

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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-\frac{1}{√x}

Than since we know x=p and y=q

Plugging in p into y' we get 1-\frac{1}{√p}

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

Giving us y - q = 1-\frac{1}{√p} (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-\frac{1}{√x}

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

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

So I did implict differention of \sqrt{x} + \sqrt{y}= 1

and get \frac{dy}{dx}=\frac{-\sqrt{y}}{\sqrt{x}}

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?
 
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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 \frac{-\sqrt{y}}{x}.

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

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

m= \frac{-\sqrt{p}}{\sqrt{q}}

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

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

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


I think this is correct?
 
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