Question

Find the equation of the tangent line to the following curve at the indicated point (20, 5).

y^2=x^2/xy-84.
y(x)=?

Answer 1

find slope dy/dx and use y-y1=slope(x-x1)

Answer 2

would i find the derivative?

Answer 3

yes

Answer 4

y/2y^3+x would this be it?

Answer 5

have you taken calc I

Answer 6

taking it

Answer 7

I dont understand this thou and the hw is due soon so im trying to finish it.

Answer 8

Use implicit differentiation to find dy/dx of the given function. Substitute the x-value into the derrivative to find the gradient/slope of the tangent. Then use y-5=m(x-20) to find the equation of the tangent. m=gradient

Answer 9

y/2y^3+x is this it?

Answer 10

\[y^2 = \frac {x^2}{xy-84}\]\[xy^3 - 84y = x^2\]Now use implicit differentiation\[\frac {d}{dx} \left[ xy^3 - 84y = x^2 \right]\]\[\frac {d}{dx} \left[ xy^3 \right] - 84 \frac {dy}{dx} = 2x\]\[y^3 + 3xy^2 \frac {dy}{dx} - 84 \frac {dy}{dx} = 2x\]\[\frac {dy}{dx} (3xy^2 - 84) = 2x - y^3\]\[\frac {dy}{dx} = \frac {2x - y^3}{3xy^2 - 84}\]

Answer 11

To do the solution I need to understand your question clearly. is the function y^2=x^2/(xy-84) or its y^2=x^2/(xy) -84 ?

Answer 12

Rogue's simplification of the given function is incorrect. Should have been xy^3 - 84y^2 = x^2. the error is on 84y, it should be 84y^2 i:e y should be squared

Answer 13

:3 damn, those little mistakes...

Answer 14

-y^3+2x/3y(xy-56)

Answer 15

that good?

Answer 16

\[\frac {d}{dx} \left[ xy^3 - 84y^2 = x^2 \right]\]\[y^3 + 3xy^2 \frac {dy}{dx} - 168y \frac {dy}{dx} = 2x\]\[\frac {dy}{dx}(3xy^2 - 168y) = 2x - y^3\]\[\frac {dy}{dx} = \frac {2x - y^3}{3y(xy - 54)}\]

Answer 17

Actually, mansukh's thing is more right, lol, I made a mistake dividing 168 by 3... epic fail.

Answer 18

\[\frac {dy}{dx} = \frac {2x - y^3}{3y(xy - 56)}\]

Answer 19

lol yeah... now for the next step Mansukh, you need to plug in your given x and y values into the derrivative equation to get the gradient of the tangent

Answer 20

then after that you can use straight line equation y=mx+c to get an equation of the tangent

Answer 21

the one that rouge put up?
2(20)-5^3/3(5)((20)(5)-56)=-9126.6666

Answer 22

Its actually -0.1287878788. Maybe you made an error on your calculator. Can you re-check

Answer 23

this is what i got 7.57576-0.128788 x i put it into the site and it wasnt rite =(

Answer 24

the questions asking for what dy/dx=

Answer 25

Are you ok now, or you still do not understand?

Answer 26

it says the answers wrong

Answer 27

There's gotta be something wrong in the given equation because the procedure and the final answer is correct. I just double checked now

Answer 28

oh well =(

Answer 29

\[Y_{tangent} = \frac {-17}{132}(x-20) + 5 = \frac {250}{33} - \frac {17x}{132} \approx 7.576 - 0.129x\]Da website sucks.