Question

Back again! want to make sure i didnt make an error here
Been a while since i did implicit diff, but here it is
Given the relationship x^3 +2y^2 =10, with y>0 and dy/dt=4 units/min., find the value of dx/dt at the instant y=1 unit.

its a nice combo of implicit and Related rates, for my final answer i got -4/3 units/min would that be correct? my methodolgy:
3x^2+4y dy/dx =0
4y dy/dx=-3x^2
4*1*4=-3x^2
SQRT16=SQRT-3x^2
4=-3x
4/-3=x--->-4/3=x is that good?

Answer 1

dy/dt * dx/dy = dx/dt did you do this step?

Answer 2

whoops no i did not

Answer 3

so would it be 3x^2dx/dy?

Answer 4

yes that works with 4y

Answer 5

you might want to backsub y=1 into the original equation for x, then you can get dx/dy value once you have both x and y values since dx/dy i got was -4y/3x^2

Answer 6

so 3x^2 dx/dy +4y dy/dt

Answer 7

then times that by dy/dt

Answer 8

3x^2 dx/dy + 4y = 0

Answer 9

I thought we got a dy/dx from deriving 4y?

Answer 10

and where do i put 4?

Answer 11

think i still missed a step

Answer 12

oh up to you if you want to solve for dy/dx or dx/dy, just have to reciprocate the dy/dx later on if you solve for dy/dx since it's dx/dy times dy/dt that equals dx/dt

Answer 13

your original 3x^2 + 4y dy/dx = 0 is correct in itself

Answer 14

so in otherwords, it wont reciprocate anything in my equation, its just something i need to be careful of in more advanced equations, something i overlooked but got lucky that it didnt effect anytrhing?

Answer 15

that dy/dx in orig post was a mistake i meant for it to be dy/dt BTW that probably means i do need to do some reciprocating

Answer 16

your answer however, is correct. it is -4/3

Answer 17

at your third line, i just would have done: dy/dx = -3x^2/4y and from subbing y=1 into x^3 + 2y^2 = 10, I get x= 2, then sub x and y = 2 and 1 into -3x^2/4y to get dy/dx, then we do 1/that, and times that by 4

Answer 18

Ok thank you for your help

Answer 19

Pooja do you have input?

Answer 20

ah ok then