by Guest » Wed Nov 06, 2019 9:16 am
It really annoys me that we cannot "edit" a post on this forum!
Guest wrote:Yes, the rate at which the water is rising is dh/dt. However, "[tex]V= \frac{1}{3}\pi r^2h[/tex]" is not enough. You also need to know how "h" varies with "r". [tex]\frac{dV}{dh}= \frac{1}{3}\pi\left(2rh\frac{dr}{dt}+ r^2\frac{dh}{dt}\right)[/tex]. In this problem you are told that "The vessel has a height of 2m and a diameter of 2m" which means that whatever the height of water in the vessel is, the radius is equal to it: r= h. We [tex]\frac{dr}{dt}= \frac{dh}{dt}[/tex]. Making those changes in the derivative, [tex]\frac{dV}{dt}= \frac{1}{3}\pi\left(2h^2\frac{dh}{dt}+ r^2\frac{dh}{dt}= \frac{1}{3}\pi )(3h^2)\frac{dh}{dt}\right)= \pi h^2 \frac{dh}{dt}[/tex]. We could also have done this by setting r= h initially: [tex]V= \frac{1}{3}\pi h^3[/tex] so [tex]\frac{dV}{dt}= \pi h^2\frac{dh}{dt}[/tex].
We are told that water is coming in at 1 cubic meter per second so [tex]\frac{dV}{dt}= 1[/tex]. We are asked how the level of the water is increasing, [tex]\frac{dh}{dt}[/tex], when the cone is 1/8 full. The full volume is [tex]\frac{1}{3}\pi (2)^3= frac{8}{3}\pi[/tex] and 1/8 of that is [tex]\frac{1}{3}\pi[/tex] cubic meters. Then [tex]\frac{1}{3}\pi h^3= \frac{1}{3}\pi[/tex]. h= r= 1 and we have [tex]\frac{dV}{dt}= 1= \pi (1)^2\frac{dh}{dt}[/tex] so [tex]\frac{dh}{dt}= \frac{1}{\pi}[/tex] meters per second, not [tex]\frac{\pi}{4}[/tex]. How did you get that?
For the second part, since r= h at every height, the surface of the water in the cone is given by [tex]A= \pi r^2= \pi h^2[/tex] so [tex]\frac{dA}{dt}= 2\pi h \frac{dh}{dt}[/tex].
It really annoys me that we cannot "edit" a post on this forum!
[quote="Guest"]Yes, the rate at which the water is rising is dh/dt. However, "[tex]V= \frac{1}{3}\pi r^2h[/tex]" is not enough. You also need to know how "h" varies with "r". [tex]\frac{dV}{dh}= \frac{1}{3}\pi\left(2rh\frac{dr}{dt}+ r^2\frac{dh}{dt}\right)[/tex]. In this problem you are told that "The vessel has a height of 2m and a diameter of 2m" which means that whatever the height of water in the vessel is, the radius is equal to it: r= h. We [tex]\frac{dr}{dt}= \frac{dh}{dt}[/tex]. Making those changes in the derivative, [tex]\frac{dV}{dt}= \frac{1}{3}\pi\left(2h^2\frac{dh}{dt}+ r^2\frac{dh}{dt}= \frac{1}{3}\pi )(3h^2)\frac{dh}{dt}\right)= \pi h^2 \frac{dh}{dt}[/tex]. We could also have done this by setting r= h initially: [tex]V= \frac{1}{3}\pi h^3[/tex] so [tex]\frac{dV}{dt}= \pi h^2\frac{dh}{dt}[/tex].
We are told that water is coming in at 1 cubic meter per second so [tex]\frac{dV}{dt}= 1[/tex]. We are asked how the level of the water is increasing, [tex]\frac{dh}{dt}[/tex], when the cone is 1/8 full. The full volume is [tex]\frac{1}{3}\pi (2)^3= frac{8}{3}\pi[/tex] and 1/8 of that is [tex]\frac{1}{3}\pi[/tex] cubic meters. Then [tex]\frac{1}{3}\pi h^3= \frac{1}{3}\pi[/tex]. h= r= 1 and we have [tex]\frac{dV}{dt}= 1= \pi (1)^2\frac{dh}{dt}[/tex] so [tex]\frac{dh}{dt}= \frac{1}{\pi}[/tex] meters per second, not [tex]\frac{\pi}{4}[/tex]. How did you get that?
For the second part, since r= h at every height, the surface of the water in the cone is given by [tex]A= \pi r^2= \pi h^2[/tex] so [tex]\frac{dA}{dt}= 2\pi h \frac{dh}{dt}[/tex].[/quote]
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