Question

The equation of a curve is sqrt x + sqrt y = sqrt a, where a is a positive constant.
Express dy/dx in terms of x and y.

Answer 1

@mim_3 any idea

Answer 2

@Mimi_x3

Answer 3

\[\sqrt{x}+\sqrt{y}=\sqrt{a}\]\[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0\]\[\frac{dy}{dx}=-\sqrt{\frac{y}{x}}\]

Answer 4

Pls explain your working then?

Answer 5

I m back

Answer 6

sorry for late reply lost connection

Answer 7

@Adi3 you weren't even helping me -.-

Answer 8

Looks like this is a separable differential equation.

So in terms of finding dy/dx in terms of y and x, we can take the derivative of the entire function with respect to x, and then just solve for dy/dx.

That's my take on it.

Answer 9

i always loose connection

Answer 10

whenever I trie to reply no conction

Answer 11

Implicit equation tho.

Answer 12

Oh yes, that is correct. Implicit as we are solving for \(y'\).

Answer 13

How do you work them out? ):

Answer 14

I don't know how to solve this.
ohh shoot don't know why

Answer 15

@Adi3 Get lost pls. I don't need your help.

Answer 16

\[\sf \sqrt{x} +\sqrt{y} = c\]\[d(\sqrt{x}+\sqrt{y}=c)\]\[d(\sqrt{x})+d(\sqrt{y}) = d(c)\]\[\frac{1}{2x^{1/2}} +\frac{1}{2y^{1/2}} \cdot y' = 0\]Solving for \(y'\) we get \[\frac{1}{2y^{1/2}}y' =-\frac{1}{2x^{1/2}}\]\[\boxed{y'=-\dfrac{y^{1/2}}{x^{1/2}} = -\left(\dfrac{y}{x}\right)^{1/2} = -\sqrt{\dfrac{y}{x}}} \]

Answer 17

I labeled \(\sqrt{a}\) as c, indicating that it is a constant.

Answer 18

the d is referring to dy/dx isit

Answer 19

\[d=\frac{d}{dx}\] And thus implicitly differentiating y with respect to x, we have \(\dfrac{dy}{dx}\)

Answer 20

Does that make sense?

Answer 21

YESSS. Thanks so much.

Answer 22

Awesome.