Question

if x^3+y^3=9 and dx/dy=4, find dy/dt when x=2

Answer 1

did i do it right

Answer 2

there is something wrong in your statement
upto y=1 you are correct
\[3x^2\frac{ dx }{ dt }+3y^2 \frac{ dy }{ dt }=0\]
\[\frac{ \frac{ dy }{ dt } }{ \frac{ dx }{ dt } }=-\frac{ x^2 }{ y^2 }\]
\[\frac{ dy }{ dx }=-\frac{ 2^2 }{ 1^2 }=-4\]

Answer 3

oh sorry idid wrong
\[actually ~i~should~find~\frac{ dx }{ dy }=-\frac{ y^2 }{ x^2 }=-\frac{ 1 }{ 4 }\]

Answer 4

wait where you getting -1/4 from is it not be -4 because dy/dx is 4

Answer 5

HI!

Answer 6

\[x^3+y^3=9\] right?

Answer 7

so \[3x^2+3y^2y'=0\\
y'=-\frac{x^2}{y^2}\]

Answer 8

oh doe, i just repeated what @surjithayer said word for word

Answer 9

this question has no meaning unless the relationship of y or x with t is defined

Answer 10

@misty1212 i got this question upto that point but know what we have to do