Question

#2- For the curve given by 4x2+y2=48+2xy show that
dy/dx=y-4x/y-x

#3- For the curve given by 4x2+y2=48+2xy, find the positive y coordinate given that the xcoordinate is 2.

#4- For the curve given by 4x2+y2=48+2xy, show that there is a point P with x-coordinate 2 at which the line tangent to the curve at P is horizontal.

Answer 1

1)
take the derivative of \(\large\color{black}{ 4x^2+y^2=48+2xy }\)
(don't forget the \(\large\color{black}{ dy/dx }\) for each derivative of y)

2)
Rearrange the equation in terms of \(\large\color{black}{ dy/dx }\)

Answer 2

this is for number 2.

Answer 3

can you find the derivative of \(\large\color{black}{ 4x^2+y^2=48+2xy }\) ?

Answer 4

i only need #4 i figured out the others :) thank u so much but can u please help me with #4

Answer 5

@SolomonZelman

Answer 6

so you want to show that the horizontal tangent of \(\large\color{black}{ 4x^2+y^2=48+2xy }\) is at x=2.

Answer 7

firstly, you need the derivative (I disconnected sorry)

Answer 8

can you tell me what will the \(\large\color{black}{ dy/dx }\) be?

Answer 9

can u show me all the steps? that's how i learn math & i'll understand through all of the steps :)

Answer 10

sure

Answer 11

~ find the derivative
~ solve for the derivative
~ the derivative of the function is it's slope, so set y' (or dy/dx, whatever notation is) =0
~ if you get x=2 after you set y'=0, then your statement in number 4 is true.

Answer 12

so can you perform the steps with me now?

Answer 13

okay, when I differentiate, I get (using y' as my notation)
\(\large\color{black}{ 8x+2y{\tiny~}y'=2y+2x{\tiny~}y' }\)

\(\large\color{black}{ 2y{\tiny~}y'-2x{\tiny~}y'=2y-8x }\)

\(\large\color{black}{{\tiny~}y'(2y-2x)=2y-8x }\)

\(\large\color{black}{{\tiny~}y'=\frac{\LARGE 2y-8x }{\LARGE 2y-2x} }\)

\(\large\color{black}{{\tiny~}y'=\frac{\LARGE y-4x }{\LARGE y-x} }\)

Answer 14

yes, so what the derivative is said to be in your question, is correct.

Answer 15

then:
\(\large\color{black}{{\tiny~}0=\frac{\LARGE y-4x }{\LARGE y-x} }\)

Answer 16

you need f(2) for this, to plug in the point (2,f(2))

Answer 17

\(\large\color{black}{4x^2+y^2=48+2xy }\)
\(\large\color{black}{4(2)^2+y^2=48+2(2)y }\)
\(\large\color{black}{16+y^2=48+4y }\)
\(\large\color{black}{y^2-4y=32 }\)
\(\large\color{black}{y^2-4y+4=32+4 }\)
\(\large\color{black}{(y-2)^2=36 }\)
\(\large\color{black}{y=2\pm6 }\)
\(\large\color{black}{y=8,~-4 }\)

Answer 18

so there are actually 2 horizontal tangents, y=8 and the other y=-4.

Answer 19

\(\large\color{black}{{\tiny~}y'=\frac{\LARGE y-4x }{\LARGE y-x} }\)

\(\large\color{black}{{\tiny~}0=\frac{\LARGE (8)-4(2) }{\LARGE (8)-(2)} }\)

\(\large\color{black}{{\tiny~}0=\frac{\LARGE 0 }{\LARGE 6} }\)

Answer 20

yes there is a horizontal tangent to the curve where x=2, and it is when y=8.

Answer 21

the graph:
https://www.desmos.com/calculator/wwlyruii03

Answer 22

I will goof around in an English section for a bit, btu I am out in 10 minutes or so

Answer 23

yeah bye, I got to go in a minute and a half

Answer 24

thank u so much!!!!!!!!!!!!!

Answer 25

u went over all of the problem perfectly! i understand it now :) ur awsum thank u