Where Is the Error in Finding the Maximum Point of y = x^2√(1 − x^2)?

[jinx007]

The curve y = x^2√(1 − x^2) for x ≥ 0 and its maximum point M. Find the coordinates of M (x coordinates) ^ --> means power

dy/dx is zero at stationary point

so i make use of the chain rule

and after simplification i get my answer as

x^3 = 2 x

so x^2 = 2 and x = square root of 2

but unfortunately the answer in the marking scheme
is square root 2/3

How is it so, please help me, where is the error...!

 

[rl.bhat]

The curve y = x^2√(1 − x^2) for x ≥ 0 and its maximum point M. Find the coordinates of M (x coordinates) ^ --> means power

dy/dx is zero at stationary point

How is it so, please help me, where is the error...!

y = x^2sqrt{(1-x^2)}

\frac{dy}{dx} = x^2\frac{-2x}{2\sqrt{(1-x^2)}} + \sqrt{(1-x^2)}2x = 0

\frac{-x^3}{\sqrt{(1-x^2)}} + \sqrt{(1-x^2)}2x = 0

Now simplify.

 

[HallsofIvy]

The curve y = x^2√(1 − x^2) for x ≥ 0 and its maximum point M. Find the coordinates of M (x coordinates) ^ --> means power

dy/dx is zero at stationary point

so i make use of the chain rule

and after simplification i get my answer as

x^3 = 2 x

We can't tell you where your error is until you tell us how you got this! Your error is somewhere in that calculation.

so x^2 = 2 and x = square root of 2

but unfortunately the answer in the marking scheme
is square root 2/3

How is it so, please help me, where is the error...!

 

[Susanne217]

y = x^2sqrt{(1-x^2)}

\frac{dy}{dx} = x^2\frac{-2x}{2\sqrt{(1-x^2)}} + \sqrt{(1-x^2)}2x = 0

\frac{-x^3}{\sqrt{(1-x^2)}} + \sqrt{(1-x^2)}2x = 0

Now simplify.

There is a mistake cause I get

y = x^2\sqrt{(1-x^2)}

\frac{dy}{dx} = \frac{-3\cdot x^3-2 \cdot x}{\sqrt{(1-x)} \cdot \sqrt{(1+x)}}

and then use the "First derivative test" shown here

http://www.math.hmc.edu/calculus/tutorials/extrema/

to find the maximum point of the function

 

[Mentallic]

There is a mistake cause I get

y = x^2\sqrt{(1-x^2)}

\frac{dy}{dx} = \frac{-3\cdot x^3-2 \cdot x}{\sqrt{(1-x)} \cdot \sqrt{(1+x)}}

and then use the "First derivative test" shown here

http://www.math.hmc.edu/calculus/tutorials/extrema/

to find the maximum point of the function

What you have is wrong. Show us what you did and we'll steer you in the right direction. The key here is to notice that you need to use the product rule.

 

[Susanne217]

What you have is wrong. Show us what you did and we'll steer you in the right direction. The key here is to notice that you need to use the product rule.

Dude according to Wolfram integrator my result is correct..

 

[Mentallic]

There is a mistake cause I get

Dude according to Wolfram integrator my result is correct..

So you didn't actually calculate the derivative yourself? Because with my understanding of derivatives, what rl.bhat got is correct and what you have is not what he has... calculator or not...

 

[Susanne217]

So you didn't actually calculate the derivative yourself? Because with my understanding of derivatives, what rl.bhat got is correct and what you have is not what he has... calculator or not...

I did calculate it and get the result above !

 

[Mentallic]

Then post what you did so I can point at where you've gone wrong.

 

[Susanne217]

Then post what you did so I can point at where you've gone wrong.

you got me pappa

after retrying the calculation I get

f'(x) = 2x\cdot \sqrt{(1-x^2)} + x^2 \cdot \frac{-2x}{2\cdot \sqrt{1-x^2}}

but still the first derivative test is the way to go to find the maximum of f..

 

[Mentallic]

There you go

 

[jinx007]

There you go

hey its so strange, i got the answer that is square root 2/3..

I make use of the dy/dx = v du - u dv / v^2

so i put x^2 / (1-x^2)^-1/2 and i solve...woOo so puzzling.. anyway thanks to all who helped me..!

 

[Mentallic]

It's not puzzling at all that you apply a different rule of differentiation to solve the same function! If it were different, there are obvious flaws in our "rules".

It works for anything you like, say we wanted to differentiate x², well we use the obvious rule to turn it into 2x, but say we instead used the quotient rule on \frac{1}{x^{-2}} then you would find the same answer, even though more work is required

But be wary, while it may not seem like the same answer at first sight, they are still equivalent so you would need to simplify/re-arrange in order to show they're equal.

 

[HallsofIvy]

You could also have done that as
y= x^2(1- x^2)^{1/2}
with the chain rule and product rule.

 

[Gregg]

Another way that works is

y^2=x^4-x^6

2yy' = 4x^3-6x^5

y' = \frac{4x^3-6x^5}{2\sqrt{x^4-x^6}}

of course y'=0 implies

4x^3-6x^5 = 0

x = \sqrt{2}/3 works