Question

All edges of a cube are expanding at a rate of 6 cm/sec. How fast is the volume changing when each edge is (a) 2 cm and (b) 10 cm?

Answer 1

a lovely related rate problem for sure.

Answer 2

\[V=x^3\] so you know
\[V'=3x^2x'\] and you are told that
\[x'=6\] so you have
\[V'=18x^2\] now replace x by your two values to get the answer

Answer 3

im confused....how did u get v'.

Answer 4

i just wrote it. in other words we know
\[V(x(t))=x^3(t)\]

Answer 5

arent we suppsed to differentiate....liek write dr/dt = smthng adn then dv/dt = smthng? thats how we didi it in class but idk how to do it 4 spheres

Answer 6

we can write it like that if you like. instead of writing
\[\frac{dx}{dt}\] i just wrote
\[x'\]

Answer 7

you can write
\[V=x^3\]
\[\frac{dV}{dt}=3x^2\frac{dx}{dt}\] if you prefer

Answer 8

still confused?

Answer 9

maybe in class you wrote
\[\frac{dV}{dt}=\frac{dV}{dx}\times \frac{dx}{dt}\]

Answer 10

i understand it this way...wat are teh rest of teh steps wen we wrtie it this way in terms of dv//dt...

Answer 11

just as i wrote it. you are told that
\[\frac{dx}{dt}=6\] in the statement "All edges of a cube are expanding at a rate of 6 cm/sec."

Answer 12

and you know if
\[V=x^3\] then
\[\frac{dV}{dx}=3x^2\]

Answer 13

put that together and you get
\[\frac{dV}{dt}=\frac{dV}{dx}\times \frac{dx}{dt}=3x^2\times 6=18x^2\]

Answer 14

so now we plug 2 and 10 into it to get the ansewrs?

Answer 15

there is a second part to teh question....can u help?

Answer 16

determine how fast the surface area is changing when each edge is a) 2 and b) 10 cm.

Answer 17

yes you plug in the numbers to get the answer

Answer 18

surface area of a cube is
\[S=6x^2\] since there are six sides each with area
\[x^2\] repeat the process and get
\[S'=12xx'\] or
\[S'=72x\]

Answer 19

since u explained this well may u help me w/ another one? this one is realll confusing me b.c it is a cone

Answer 20

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 10 cubic ft/ min. The diameter of the base of the cone is approximately 3x the altitude. At what rate is the height of the pile changing when the pile is 15 ft high?

Answer 21

so what formula do we use to plug in 2 and 10? is it V=3x^2dx/dt?