Question

the equation of a curve is (x)^1/2 + (y)^1/2=(a)^1/2
where a is a positive constant
i) express dy/dx in terms of x and y.
ii) the straight line with equation y=x intersects the curve at the point P. find the equation of the tangent to the curve at P.

Answer 1

i)\[d/dx(\sqrt{x}+\sqrt{y})=d/dx \sqrt{a}\]
\[1/2\sqrt{x}+(1/2\sqrt{y}\times dy/dx0)=0\]
\[-1/2\sqrt{x}\div 1/2\sqrt{y}=-\sqrt{(y/x)}=dy/dx\]
ii) \[1)y=x\]\[2) y=(\sqrt{a}-\sqrt{x})^2\]

Answer 2

\[x=(\sqrt{a}-\sqrt{x})^2\rightarrow 2\sqrt{x}=\sqrt{a}\rightarrow x=a/2\]

Answer 3

\[y=a/2 \] i am not sure about what y equals to.

Answer 4

\[dy/dx=-\sqrt{(a/2)/(a/2)}=-1\]

Answer 5

\[y-y1=m(x-x1)\rightarrow y-a/2=-1(x-a/2)\rightarrow y+x=a\]

Answer 6

so it looks like you differentiate your curve equation correctly:
\[\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} y'=0 \\ \frac{1}{2 \sqrt{y}} y'=-\frac{1}{2 \sqrt{x}} \\ y'=-\frac{2 \sqrt{y}}{2 \sqrt{x}}=-\frac{\sqrt{y}}{\sqrt{x}}\]
So your y' is correct.

Answer 7

Now the second part tells us we need to first first the intersection of y=x and the curve given.
Then find the tangent line at that point.

Answer 8

so we have y=x
I can replace y with x or x with y
But we only need to do one of those things so we can solve for either x or y first you know depending on which we replacement we do

Answer 9

\[x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2} \\ \text{ I will \choose \to replace } y \text{ with } x \\ x^\frac{1}{2}+x^\frac{1}{2}=a^\frac{1}{2} \\ 2 x^\frac{1}{2}=a^\frac{1}{2} \\ 4x=a \\ x=\frac{a}{4} \\ \text{ and since } y=x \\ \\ \text{ then the point of intersection is } (\frac{a}{4}, \frac{a}{4}) \]
by the way a should be positive so our point should be in the 1st quadrant
but whatever I don't even think that observation will be needed

Answer 10

So anyways you need to find the tangent line at (a/4,a/4)

Answer 11

Oh yeah I see what you did you...
you got to \[2 x^\frac{1}{2}=a^\frac{1}{2} \\ \text{ squaring both sides gives } (2 x^\frac{1}{2})^2=(a^\frac{1}{2})^2 \\ (2)^2(x^\frac{1}{2})^2=(a^\frac{1}{2})^2 \\ 4x=a\]
I hope this part is more clear

Answer 12

and your question about "I'm not sure what y equals"
well we are given y=x
so whatever we get for x that is also y

Answer 13

so you are correct to say your slope is -1
that is good
\[y-y_1=m(x-x_1) \\ y-y_1=-1(x-x_1) \\ \text{ and we also have the point we want \to find the tangent line is } (\frac{a}{4},\frac{a}{4}) \\ y-\frac{a}{4}=-1(x-\frac{a}{4})\]
I will let you put that in slope intercept form if you prefer that form

Answer 14

why did you square both sides?

Answer 15

\[2 \sqrt{x}=\sqrt{a} \\ \text{ you can't \square part of a side } \\ (2 \sqrt{x})^2=(\sqrt{a})^2 \\ \text{ and by law of exponents we can write the \left hand side as } (2)^2 (\sqrt{x})^2=\sqrt{a}\]

Answer 16

to solve for x
x had a square root around it

Answer 17

i get what you mean, thank you @freckles :D

Answer 18

With actual numbers
so you can see better maybe:
Say we have we want to solve:
\[2 \sqrt{x} =4 \\ \text{ squaring both sides gives } 4 x=16 \\ x=\frac{16}{4}=4 \\ \text{ we can \check 4} \\ 2 \sqrt{4}=2 (2)=4 \text{ \it indeed works } \\ \\ \text{ now we can try \to solve } 2 \sqrt{x}=4 \text{ squaring parting of a side } 2 x=16 \\ x=8 \\ \text{ you will see 8 will \not work }\]

Answer 19

\[2 \sqrt{x}=\sqrt{a} \\ \text{ you could have divided both sides by 2 first then \square } \\ \sqrt{x}=\frac{\sqrt{a}}{2} \\ x=(\frac{\sqrt{a}}{2})^2=\frac{a}{4}\]

Answer 20

no problem