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r(t) === will be the radius of your cone. This radius various in function of time (thats why I put (t))
V=(Pi*(r^2)*h)/3=(Pi*(r^2)*(3r/4))/3=(Pi*(r^3))/4 === Thats the volume of your cone (h=3d/8=3r/4).
V(t)=(Pi*(r(t)^3))/4 ===(1) Your volume and radius are both various in vonction of time.
r(0)=8/3 m === Your initial radius
V(0)=(Pi*((8/3)^3))/4= (128*Pi)/27 === Your initial volume
V(t)=V(0)+V.dot= (128*Pi)/27 +15*t ===(2) This is the equation of your volume in function of time
Take equations (1) and (2) togetter:
(128*Pi)/27 +15*t=(Pi*(r(t)^3))/4 === From this point all you need is to isolate your function r(t)
r(t)^3=512/27+60t/Pi
r(t)=(512/27+60t/Pi)^1/3 === You now have to derive this then its kidna over
To derive this use the substitution
u(t)=512/27+60t/Pi === Remember that: du/dt=60/Pi ...... and dt/du=Pi/60 (we'll use that later)
You'll have:
r(u)=u^1/3
dr/du=(1/3)*(u^-2/3) === Remember that: dr/du=(dr/dt)*(dt/du) and you will find:
(dr/dt)*(Pi/60)==(1/3)*(u^-2/3)=(1/3)*((512/27+60t/Pi)^-2/3)
dr/dt=(20/Pi)*((512/27+60t/Pi)^-2/3)
Thats what we were looking for!! They asked for the speed when h=2 meters... and when h=2 meters r=8/3 m thats when time t=0...
dr/dt=(20/Pi)*((512/27+0)^-2/3)=0,1111905573 m/min=11,12 cm/min
if you want to know the rate after 5min t=5 min, plug it in
r(t) === will be the radius of your cone. This radius various in function of time (thats why I put (t))
V=(Pi*(r^2)*h)/3=(Pi*(r^2)*(3r/4))/3=(Pi*(r^3))/4 === Thats the volume of your cone (h=3d/8=3r/4).
V(t)=(Pi*(r(t)^3))/4 ===(1) Your volume and radius are both various in vonction of time.
r(0)=8/3 m === Your initial radius
V(0)=(Pi*((8/3)^3))/4= (128*Pi)/27 === Your initial volume
V(t)=V(0)+V.dot= (128*Pi)/27 +15*t ===(2) This is the equation of your volume in function of time
Take equations (1) and (2) togetter:
(128*Pi)/27 +15*t=(Pi*(r(t)^3))/4 === From this point all you need is to isolate your function r(t)
r(t)^3=512/27+60t/Pi
r(t)=(512/27+60t/Pi)^1/3 === You now have to derive this then its kidna over
To derive this use the substitution
u(t)=512/27+60t/Pi === Remember that: du/dt=60/Pi ...... and dt/du=Pi/60 (we'll use that later)
You'll have:
r(u)=u^1/3
dr/du=(1/3)*(u^-2/3) === Remember that: dr/du=(dr/dt)*(dt/du) and you will find:
(dr/dt)*(Pi/60)==(1/3)*(u^-2/3)=(1/3)*((512/27+60t/Pi)^-2/3)
dr/dt=(20/Pi)*((512/27+60t/Pi)^-2/3)
Thats what we were looking for!! They asked for the speed when h=2 meters... and when h=2 meters r=8/3 m thats when time t=0...
dr/dt=(20/Pi)*((512/27+0)^-2/3)=0,1111905573 m/min=11,12 cm/min
if you want to know the rate after 5min t=5 min and just plug your time in dr/dt=...
haha, guess math isnt your thing either. No ferrari for you as well
whats wrong with nerds?
fair enough, but who do you mean by "us" ? Are you hoping to work at the geek squad at best buy or something?
