question\nwater is draining from the bottom of a cone - shaped funnel at the rate of 0.8ft³/sec. the height…

question\nwater is draining from the bottom of a cone - shaped funnel at the rate of 0.8ft³/sec. the height of the funnel is 6ft and the radius at the top of the funnel is 5ft, as shown in the image below. at what rate is the height of the water in the funnel changing when the height of the water is 4ft?\nsubmit an exact answer in terms of π. recall that the formula for the volume of a cone is\nv = 1/3πr²h.\nprovide your answer below.\ndh/dt = □ft/sec\nfeedback more instruction show steps submit\ncontent attribution

question\nwater is draining from the bottom of a cone - shaped funnel at the rate of 0.8ft³/sec. the height of the funnel is 6ft and the radius at the top of the funnel is 5ft, as shown in the image below. at what rate is the height of the water in the funnel changing when the height of the water is 4ft?\nsubmit an exact answer in terms of π. recall that the formula for the volume of a cone is\nv = 1/3πr²h.\nprovide your answer below.\ndh/dt = □ft/sec\nfeedback more instruction show steps submit\ncontent attribution

Answer

Explanation:

Step1: Relate radius and height

Since the large - cone and the cone of water are similar, we have the ratio $\frac{r}{h}=\frac{5}{6}$, so $r = \frac{5}{6}h$.

Step2: Substitute r into volume formula

Substitute $r=\frac{5}{6}h$ into $V=\frac{1}{3}\pi r^{2}h$. Then $V=\frac{1}{3}\pi(\frac{5}{6}h)^{2}h=\frac{25}{108}\pi h^{3}$.

Step3: Differentiate with respect to time

Differentiate $V$ with respect to $t$ using the chain - rule. $\frac{dV}{dt}=\frac{25}{36}\pi h^{2}\frac{dh}{dt}$.

Step4: Substitute known values

We know that $\frac{dV}{dt}=- 0.8$ (negative because the volume is decreasing) and $h = 4$. Substitute these values into $\frac{dV}{dt}=\frac{25}{36}\pi h^{2}\frac{dh}{dt}$: $-0.8=\frac{25}{36}\pi\times4^{2}\times\frac{dh}{dt}$. $-0.8=\frac{25}{36}\pi\times16\times\frac{dh}{dt}$. $-0.8=\frac{400}{36}\pi\times\frac{dh}{dt}$. $\frac{dh}{dt}=\frac{-0.8\times36}{400\pi}=-\frac{28.8}{400\pi}=-\frac{9}{125\pi}$ ft/sec.

Answer:

$-\frac{9}{125\pi}$