Question

The equation of a curve is y = √(8x − x2). Find
(i) an expression for
dy
dx
, and the coordinates of the stationary point on the curve,

Answer 1

i get x=0 ,x=-4 , x=8, x=4

Answer 2

Did you get the following for dy/dx?
\[dy/dx=1/2(8x-x ^{2})^{-1/2}\times (8-2x)\]

Answer 3

\[y=(\sqrt{8x-x^2})\]
\[y=(8x-x^2)^{\frac{1}{2}}\]
\[y=\frac{1}{2}(8x-x^2)^{-1/2}(8-2x)\]
\[4(8x-x^2)^{-1/2}-x(8x-x^2)^{-1/2}=0\]
\[4(8x-x^2)^{-1/2}=x(8x-x^2)^{-1/2}\]
\[\frac{4}{\sqrt{8x-x^2}}=\frac{x}{\sqrt{8x-x^2}}\]
\[\frac{16}{8x-x^2}=\frac{x^2}{8x-x^2}\]
\[128x-16x^2=8x^3-x^4\]
\[x^4-8x^3-16x^2+128x=0\]
\[x(x^3-8x^2-16x+128)=0\]
x=0, x=-4,x=8,x=4

Answer 4

ans is x=0, x=8

Answer 5

I get 2 extra values?

Answer 6

The coordinates of a stationary point on the curve should be an (x, y) point.

Answer 7

@Calculator , The mistake you are doing is by squaring the following
\[\frac{4}{\sqrt{8x-x^2}}=\frac{x}{\sqrt{8x-x^2}}\]

By squaring, you are increasing the number of solutions

Answer 8

how u solve it if dont square

Answer 9

From your solutions, you can eliminate x = -4 because
x(8−x)>=0

Answer 10

Now why x = 4 should not be your solution is the thing I have to investigate. Hope you understood why x = -4 cannot be your solution

Answer 11

\[y'=\frac{1}{2}\frac{1}{(8x-x^2)}(8-2x) = 0\]
\[y'=\frac{(4-x)}{ \sqrt {(8x-x^2)}} = 0\]
So according to me
\[x \neq 0,8\]
Because if we substitute x=0 or x=8, then y' becomes undeined as denominator becomes zero

Answer 12

*undefined

Answer 13

y^2 = 8x-x^2
2y dy = 8-2x dx
dy/dx = (4-x)/y zero when x = 4 (so y = 4)

Answer 14

@estudier , but calculator is saying his solution is x=0,8.. How is it possible. Even I am getting x=4

Answer 15

See this comment by @calculator

"ans is x=0, x=8"
"I get 2 extra values?"

Answer 16

He appears to be treating the equation as an ellipse but equation is only valid for positive x.
(The 0 values are minima not maxima)

Answer 17

Well, I know this is circle equation. But he says the solution given to him is x=0,8 which is puzzling me. According to me x=4 is the only solution to this. Maybe he must be having the wrong solution . Don't know .....

Answer 18

At 0 and 8 he has minima, not maxima.

Answer 19

And only if you consider as ellipse which is not valid

Answer 20

@estudier , you are right that x=0,8 is the point where minima occurs.
And maxima occurs at x=4


Now,The stationary point is where the gradient has flatlined,and that point is anywhere the derivative = 0 .

So the answer for the question where stationary point is present is x=4, right??

Answer 21

Yes

Answer 22

Ok. :D