Question

if y^3=16x^2 determine dx/dt when x=4 and dy/dt=1

Answer 1

i got dy/dx= 32x/3y^2, not sure if thats right

Answer 2

seems cool to me

Answer 3

\[\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt}}\]

Answer 4

\[\frac{\partial y}{\partial t} \cdot \frac{\partial y}{\partial x} =\frac{\partial y}{\partial x}\cdot \frac{\partial x}{\partial t}\]

Answer 5

h yeah, same kind of notation.

Answer 6

\[\frac{dx}{dt} \frac{dy}{dx}=\frac{dy}{dt} \\ \text{solve for } \frac{dx}{dt}\]

Answer 7

so what would i do for the next step?

Answer 8

Oh you know what, let's figure out y first.

Answer 9

Because that is the variable that is missing.

Answer 10

\[y^3 = 16x^2~ , ~ x=4\]\[y^3=16 \cdot (4)^2\]\[y^3 = 2^4 \cdot 2^4\]\[y^3 = 2^8\]\[y=\sqrt[3]{2^8}=2^{8/3}\]

Answer 11

Are you with me so far?

Answer 12

yea

Answer 13

now we apply this:\[\frac{\partial y}{\partial t} \cdot \frac{\partial y}{\partial x} =\frac{\partial y}{\partial x}\cdot \frac{\partial x}{\partial t}\]

Answer 14

wait, i got 256^1/3

Answer 15

so : \[3y^2 \cdot y' = 32x \cdot x'\]Now just solve for x'.\[x' = \frac{3y^2y'}{32x}\]\[x'= \frac{3\left(2^{8/3}\right)^2 \cdot (1)}{32(4)}\]

Answer 16

i got .118118, ?

Answer 17

\[x'= \frac{3(32(2^{1/3}))}{128} = \frac{96(2^{1/3})}{128} = \frac{3(2^{1/3})}{4}\]

Answer 18

got it, its .944941

Answer 19

Is that one of your answer choices?

Answer 20

thank you soo much :)

Answer 21

yea

Answer 22

Awesome!! :)

Answer 23

Oh, on a side now. How to simplify \(2^{8/3})^2\)

Answer 24

Make sure your base is \(\ge 0\)

Answer 25

Then multiply your outer power \(2\) into the power of your base: \(8/3\)

Answer 26

That becomes : \(2^{16/3}\)

Answer 27

then 32(2^(1/3))

Answer 28

Uhh... then by thinking of how you can separate \(16/3\) into an easier,simpler fraction, You can separate it like so: \(2^{15/3} \cdot 2^{1/3}\).

15/3 can be simpplied to 5. \(\implies 2^5 \cdot 2^{1/3}\)

Answer 29

:) Hope that helps!

Answer 30

yeah it does, a lot lol thanks

Answer 31

pulling an all nighter for finals lol smh

Answer 32

Yeah same here.

Answer 33

Supposed t be studying Chemistry, but I would rather study math. Haha.

Answer 34

haha im suppose to be reviewing philosophy but math is my main focus

Answer 35

/highfive! Good luck with your exams :)

Answer 36

you too :)