Find the slope of the tangent line to the curve \sqrt{1x +4y} + \sqrt{4xy} = 17 at the point ( 7,5 ). So far I have (1/2)(x+4y)^(-1/2)(1+4y')+(1/2)(4xy)^(-1/2)(4y)+(4xy')=0 but I don't know what to do next.

Question

Find the slope of the tangent line to the curve \sqrt{1x +4y} + \sqrt{4xy} = 17 at the point ( 7,5 ).        So far I have (1/2)(x+4y)^(-1/2)(1+4y')+(1/2)(4xy)^(-1/2)(4y)+(4xy')=0 but I don't know what to do next.

terenzreignz · Answer

$$\large\sqrt{x +4y} + \sqrt{4xy} = 17$$

terenzreignz · Answer

You differentiated implicitly?

anonymous · Answer

ah yes, sry for the delay, I placed y' where dy/dx should be

terenzreignz · Answer

It doesn't really matter.  Now, once you've differentiated implicitly, you can solve for y', right?

anonymous · Answer

that's the problem actually, the algebra is really "funky" and I keep getting the wrong answer :(, would you like me to type what I have so far?

terenzreignz · Answer

If you're willing.

anonymous · Answer

Indeed, it will take a bit of time though, sry about the wait

anonymous · Answer

(1/2)(x+4y)^(-1/2)(1)+(1/2)(x+4y)(-1/2)(4xy')=(-1/2)(x+4y)^(-1/2)-(4xy)^(-1/2)(4y)

anonymous · Answer

The next step was to take the common y' out, which I will continue to type

anonymous · Answer

1/2(4y')[(x+4y)^(-1/2)+(4xy)^(-1/2)(x)] = (-1/2)[(x+4y)^(-1/2)+(4xy)^(-1/2)(4y)

anonymous · Answer

that 1/2 in the beginning is parenthesized

anonymous · Answer

continuing, I divide out the equation with the equation on the left side to get

anonymous · Answer

y'= -[(x+4y)^(-1/2)-(4xy)^(-1/2)(4y)]/4[(x+4y)^(-1/2)+(4xy)^(-1/2)(x)

anonymous · Answer

finally I plug in the values for x and y to get the slope; but my application says it is incorrect

terenzreignz · Answer

You did this step?

terenzreignz · Answer

$$\huge \frac{1+4y'}{2\sqrt{x+4y}}+\frac{xy'+y}{4\sqrt{xy}}=0$$Missed a spot ^.^

anonymous · Answer

Could you show me where after I attach the file?  I see it's being rationalized, but I can't see where.

terenzreignz · Answer

I'm not good at debugging other people's work, I can barely debug my own :D

anonymous · Answer

ah, is the untitled.png file too sloppy?

terenzreignz · Answer

Maybe ^.^
But you agree with this step, right?
$$\huge \frac{1+4y'}{2\sqrt{x+4y}}+\frac{xy'+y}{4\sqrt{xy}}=0$$

terenzreignz · Answer

No need to rationalize.  In fact, you can solve for y' now, just plug in x and y, and solve away :)

anonymous · Answer

i will try and let you know, thanks for your help

anonymous · Answer

The work on paint, if it's easier to see.

anonymous · Answer

I'm uncertain in how to isolate the y' :(

terenzreignz · Answer

you don't have to, at least not yet, you can just substitute 7 for x and 5 for y, and only then isolate the y'

anonymous · Answer

here's what I get [(1+4y')/(2(27)^(1/2)]+[(7y'+5)/(4(35)^(1/2)] = 0.  I don't know where to go next since I'm not sure if I could factor the 4y' here.

terenzreignz · Answer

It's just a linear equation, isn't it?  A messed up one, but still a linear equation, with the same idea... bring everything with no y' to the other side, and only then, factor out the y'

anonymous · Answer

Ya lol, that's exactly the problem, I've been doing so much with calculus that I'm having big problems with the basics.

terenzreignz · Answer

Well, review, then :D

anonymous · Answer

i wish i knew where to start, im looking through linear equations but they aren't giving me a similar equation

anonymous · Answer

do i need to take the conjugate?

terenzreignz · Answer

What? No, just simple linear equation in one variable logic :)

anonymous · Answer

hmm, i keep seeing problems where they get the common denominators, but is that the route to go?

terenzreignz · Answer

No, but just consider them as fractional coefficients.

anonymous · Answer

hmm so like 2(27)^(1/2)?

anonymous · Answer

could you explain the methodology?  i don't think i can figure out the steps unfortunately

terenzreignz · Answer

$$\huge \frac{1+4y'}{2\sqrt{x+4y}}+\frac{xy'+y}{4\sqrt{xy}}=0$$
x = 7
y = 5
$$\huge \frac{1+4y'}{2\sqrt{27}}+\frac{7y'+5}{4\sqrt{35}}=0$$
$$\huge \frac{1}{6\sqrt{3}}+\frac{4}{6\sqrt{3}}y'+\frac{7}{4\sqrt{35}}y'+\frac{5}{4\sqrt{35}}=0$$

terenzreignz · Answer

That was just substitution.

terenzreignz · Answer

From here, it's just a combination of linear equations in one variable, and working with radicals...

You really need to proceed here on your own, or else begin a major review of linear equations :)

anonymous · Answer

you are right, i definitely need to review.  and to further check, after moving all the constants to one side, and factoring out the y', that would mean the fractions would be divided out with another?

anonymous · Answer

In other words, just to check, it would be division between fractions? (i just want to know if im going the right route)

terenzreignz · Answer

Yes, dividing fractions is essential.

anonymous · Answer

Btw, what was the reason you changed the 6 radical 3?  Would answers have been different? or is it just simplification?

terenzreignz · Answer

Just simplification.
$$\huge 2\sqrt{27}=2\sqrt{9\cdot3}=2\cdot\sqrt9\cdot\sqrt3=2\cdot3\sqrt3$$

anonymous · Answer

I tried it, but says this was incorrect.. 

[(-1)/((6)(3)^(1/2))-(5)/(4(35)^(1/2))]/[(4)/((6)(3)^(1/2))+(7)/((4)(35)^(1/2))]

terenzreignz · Answer

Well, I don't know if you followed the formula, but if you followed it, and still had it wrong, then I don't know how to do it T.T

anonymous · Answer

I tried it other ways too, someone else tried to plug in the x and y values near the beginning and solve it that way but it still didn't work.  Would this sorta problem actually be on a test?

terenzreignz · Answer

I would think so.  But I'd have answered it the way I said.

anonymous · Answer

Indeed, but I still can't see where you got the equation you sent me, I'll try to look at it some more.  I'm going to make another post, thanks for your help.