Differentiation Question (1 Viewer)

[shaon0]

The edges of a square are increasing at a rate of 5cm per a minute.
At what rate is the area of the square increasing, when the edge is 3cm long?
​

 

[lolokay]

dx/dt = 5/min
dx^2/dx = 2x
if x = 3
dx^2/dt = 30cm²/minute

 

[shaon0]

dx/dt = 5/min
dx^2/dx = 2x
if x = 3
dx^2/dt = 30cm²/minute

how did you get dx^2/dx=2x?
Sorry if it is a stupid question.

 

[lolokay]

just from the basic rule that d(xⁿ)/dx = nx^n-1

 

[shaon0]

just from the basic rule that d(xⁿ)/dx = nx^n-1

but u would have required an equation in the beginning to do that.
How'd you get the equation?

 

[lolokay]

I don't really get what you're asking =/. You want to find dx^2/dt, and dx/dt is given, and dx^2/dx is always 2x

 

[shaon0]

I don't really get what you're asking =/. You want to find dx^2/dt, and dx/dt is given, and dx^2/dx is always 2x

so is it always constant?

 

[shaon0]

I don't really get what you're asking =/. You want to find dx^2/dt, and dx/dt is given, and dx^2/dx is always 2x

Ok thanks a lot for your help

 

[lolokay]

yeah. it wouldn't really make sense for it to not be constant

 

[conics2008]

hey lolokay..

you should of just explained it like this

let the sides of the square =x

therefore its expanding at dx/dt=5

but area A= x^2 hence da/dx = 2x

therefore da/dx * dx/dt = da/dt = 2(3) * 5 =30

good day.

 

[u-borat]

^^yeah, conics approach is the generic approach.

 

[shaon0]

hey lolokay..

you should of just explained it like this

let the sides of the square =x

therefore its expanding at dx/dt=5

but area A= x^2 hence da/dx = 2x

therefore da/dx * dx/dt = da/dt = 2(3) * 5 =30

good day.

Oh i got it....ok.

 

[lolokay]

oh.. I forgot to state that the side lengths were 'x' and therefore area is 'x²'