Question

Find the slope of the tangent line to the curve
sqrt(4x+1y)+sqrt(4xy)=13.7 at the point (6,3)

Answer 1

@jim_thompson5910

Answer 2

the square roots really throw me off, so idk :/

Answer 3

ick
does it really say
\[\sqrt{4x+y}+\sqrt{4xy}=13.7\]?

Answer 4

yeah

Answer 5

lol your math teacher must hate you

Answer 6

yessss lol

Answer 7

before i write all the muck we gotta write, lets at least make \(\sqrt{4xy}=2\sqrt{xy}\) okay?

Answer 8

ok

Answer 9

and also recall that the derivative of \(\sqrt x\) is \(\frac{1}{2\sqrt{x}}\)

Answer 10

now just grind it out
\[\frac{4+y'}{2\sqrt{4x+y}}+\frac{y+xy'}{\sqrt{xy}}=0\]

Answer 11

probably went with too few steps
any idea how i got that?

Answer 12

okay so the derivative of \[\sqrt{4x+y}\]=4+y' then divided by 2\[\sqrt{4x+y}\]

Answer 13

right

Answer 14

the second part im kind of confused

Answer 15

ok well the twos cancelled for one

Answer 16

we pulled out the two, and that goes with the two in the denominator

Answer 17

then for the derivative of \(xy\) we need the product rule

Answer 18

oh okay, got it

Answer 19

good
so the derivative of \(xy\) wrt x is \(y+xy'\) via the product rule, and that goes in the numerator

Answer 20

now replace x by 6,y by 3 and solve for \(y'\)

Answer 21

so, \[\frac{ 4+y' }{ 6\sqrt{3} }+\frac{ 3+6y' }{ 3\sqrt{2} }=0\]?

Answer 22

would you multiply both denominators?

Answer 23

@jim_thompson5910 ?

Answer 24

yeah you could multiply both sides by the LCD \[\Large 6\sqrt{6}\] to clear out the fractions

Answer 25

sqrt(4x+y)+sqrt(4xy) is approximately 13.7 at (x=6,y=3)

Answer 26

http://www.wolframalpha.com/input/?i=sqrt%284*6%2B3%29%2Bsqrt%284*6*3%29

Answer 27

i got y'=-1, it says its wrong

Answer 28

so what I'm saying is this point is not totally on any tangent line to the curve

Answer 29

it would be really close but not on

Answer 30

now im confused lol

Answer 31

\[\frac{4+y'}{2\sqrt{4x+y}}+\frac{y+xy'}{\sqrt{xy}}=0 \\ \frac{4+y'}{2 \sqrt{4(6)+3}}+\frac{3+6y'}{\sqrt{6(3)}}=0 \\ \frac{4+y'}{2 \sqrt{27}}+\frac{3+6y'}{\sqrt{18}}=0 \\ \frac{4+y'}{2 \sqrt{9 \cdot 3}}+\frac{3+6y'}{\sqrt{9 \cdot 2}}=0 \\ \frac{4+y'}{6 \sqrt{3}}+\frac{3+6y'}{3 \sqrt{2}}=0 \\ \text{ multiply both sides by } 6 \sqrt{6} \text{ as suggested by jim } \\ 6 \sqrt{6} \frac{4+y'}{6 \sqrt{3}}+6 \sqrt{6} \frac{3+6y'}{3 \sqrt{2}}=0 \\ \sqrt{2} (4+y')+2 \sqrt{3}(3+6y')=0 \\ 4 \sqrt{2} +\sqrt{2}y' +6 \sqrt{3}+12 \sqrt{3}y'=0\]
and you think this will end up being y'=-1?
but anyways I think this question itself is a trap because as I said (6,3) is not even on the curve
it is near but not on

Answer 32

yeah i got y'=-1.19112, what do you get?

Answer 33

@freckles

Answer 34

\[\sqrt{2}y'+12 \sqrt{3}y'=-4 \sqrt{2}-6 \sqrt{3} \\ (\sqrt{2}+12 \sqrt{3})y'=-4 \sqrt{2}-6 \sqrt{3} \\ y'=\frac{-4 \sqrt{2}-6 \sqrt{3}}{\sqrt{2}+12 \sqrt{3}}\]
but as i said before finding y' here shouldn't matter
because the point (6,3) is not even on the given curve so you shouldn't be able to find the tangent to the given curve at that point ...

Answer 35

i guess my professor did a mistake then, i emailed him already

Answer 36

thanks anyways