Question

The area of the cube is increasing at a rate 4 cm^2/s. How fast is the volume of the cube increasing when the edge length is 10
cm?

Answer 1

Given: \[\frac{ dA }{ dt }=4cm ^{2}/s\]
\[V=L*W*H\]
\[\frac{ dV }{ dt }=WH*\frac{ dL }{ dt }+LH*\frac{ dW }{ dt }+LW*\frac{ dH }{ dt }\]
Since this is a cube all the sides are equal.
\[\frac{ dV }{ dt }=100(\frac{ dL }{ dt }+\frac{ dW }{ dt }+\frac{ dH }{ dt })\]
\[A=6a^2, \frac{ dA }{ dt }=12a, a=\frac{ 4 }{ 12 }\]
\[\frac{ dV }{ dt }=100(\frac{ 4 }{ 12 })=\frac{ 100 }{ 3 }cm ^{3}/s\]
Is this right?

Answer 2

is it just me or r u guys also losing connection every few minutes

Answer 3

yes I am
:(

Answer 4

ok anyways a seems to be always changing w.r.t to time
so a=a(t)
where a is an edge of a cube
\[A(t)=6 [a(t)])^2\]
\[V(t)=[a(t)]^3\]

we are given A'(t)=4

Answer 5

s = side length of cube


A = surface area of cube
A = 6s^2
dA/dt = 12s*ds/dt
4 = 12s*ds/dt
ds/dt = (4)/(12s)
ds/dt = 1/(3s)


V = volume of cube
V = s^3
dV/dt = 3s^2*(ds/dt)
dV/dt = 3s^2*(1/(3s))
dV/dt = (3s^2)/(3s)
dV/dt = s
dV/dt = 10

Answer 6

\[A'(t)=12 a(t) \cdot a'(t)\]
\[12 a(t) \cdot a'(t)=4\]
Solve for a'(t)\]

we will later use a(t) is 10 after we take care of all the differentiating

Answer 7

nevermind @jim_thompson5910 beat me to it

Answer 8

lol no worries

Answer 9

ah ok I see

Answer 10

Thank you guys!