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

  • Thread starter SteliosVas
  • Start date
In summary, the students are struggling with a problem involving points (p,q) on the curve √x + √y = 1. They have rearranged and taken the derivative, but are still having trouble. They are considering using implicit differentiation and trying other methods.
  • #1
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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  • #2
For starters, check that you have rearranged & taken the derivative correctly. Maybe doing dx/dy, as well, will get you somewhere. Just guessing!
 
  • #3
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.
 
  • #4
Something else that might help is:

[tex]p\sqrt{q} + q\sqrt{p} = \sqrt{pq}(\sqrt{p}+\sqrt{q})[/tex]
 
  • #5
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:
  • #6
Remember that ##\sqrt{p} + \sqrt{q} = 1##.
 
  • #7
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?
 
  • #8
ImageUploadedByPhysics Forums1409547175.334908.jpg


I think this is correct?
 
Last edited:

1. What is the slope of a curve at a specific point?

The slope of a curve at a specific point is a measure of the steepness of the curve at that point. It represents the rate of change of the curve at that point.

2. How do you find the slope of a curve at a specific point?

To find the slope of a curve at a specific point, you can use the derivative. The derivative is a mathematical tool that gives the slope of the curve at any point on the curve.

3. What is the significance of the slope of a curve at a specific point?

The slope of a curve at a specific point is significant because it can help us understand the behavior of the curve at that point. It can also provide information about the rate of change of the curve at that point.

4. How does the slope of a curve change at different points?

The slope of a curve can change at different points depending on the shape of the curve. For example, a curve that is steep at one point may have a smaller slope at another point.

5. Can the slope of a curve be negative?

Yes, the slope of a curve can be negative. A negative slope indicates that the curve is decreasing or getting flatter as it moves from left to right. A positive slope represents an increasing or steepening curve.

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