7. a water tank has the shape of an inverted circular cone. the height of the cone is twice of the radius…

Explanation:

Step1: Relate radius and height

For a cone, the ratio of radius to height is constant. Given height of cone $h = 7m$ and assume radius of cone $r$ at height $h$ is related as $\frac{r}{h}=\frac{R}{H}$ (where $R$ and $H$ are the base - radius and height of the whole cone). Since the cone is inverted and we know the relationship between $r$ and $h$ for similar - triangles property. Let the height of water at any time $t$ be $h$ and radius of water - surface be $r$. The volume of a cone of water $V=\frac{1}{3}\pi r^{2}h$. From similar triangles, if the height of the whole cone is $H$ and base - radius is $R$, and for our case, assume the whole cone has height $H = 7m$ and we don't know $R$ yet, but for the water - cone part, $\frac{r}{h}=\frac{R}{H}$. Let's assume the base - radius of the whole cone is $R$ and height $H = 7m$. For the water - cone, we have $\frac{r}{h}=\frac{R}{7}$. Also, we know that the volume of water $V$ is being filled at a rate $\frac{dV}{dt}=2m^{3}/min$.

Step2: Express volume in terms of one variable

Since the cone is similar, if we assume the ratio of radius to height of the whole cone is a constant. Let's assume the base - radius of the whole cone is $R$. For the water - cone, $r=\frac{R}{7}h$. The volume of water $V = \frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi(\frac{R}{7}h)^{2}h=\frac{\pi R^{2}}{147}h^{3}$. Differentiate $V$ with respect to time $t$ using the chain rule: $\frac{dV}{dt}=\frac{\pi R^{2}}{49}h^{2}\frac{dh}{dt}$. We are given $\frac{dV}{dt} = 2m^{3}/min$ and $h = 3m$. We need to find $\frac{dh}{dt}$.

Step3: Solve for $\frac{dh}{dt}$

Substitute $\frac{dV}{dt}=2$, $h = 3$ into $\frac{dV}{dt}=\frac{\pi R^{2}}{49}h^{2}\frac{dh}{dt}$. First, note that from the similar - triangles property of the whole cone, if we assume the base - radius of the whole cone is $R$ and height $H = 7$, for the water - cone part, the volume formula $V=\frac{1}{3}\pi r^{2}h$. Since $\frac{r}{h}=\frac{R}{7}$, we can also write $V=\frac{1}{3}\pi(\frac{R}{7}h)^{2}h$. Differentiating gives $\frac{dV}{dt}=\frac{\pi R^{2}}{49}h^{2}\frac{dh}{dt}$. We know $\frac{dV}{dt}=2$, $h = 3$. Then $2=\frac{\pi R^{2}}{49}\times3^{2}\times\frac{dh}{dt}$. Since for a similar cone, the ratio of radius to height is constant. Let's assume the base - radius of the whole cone is $R$ and height $H = 7$. The volume of water $V=\frac{1}{3}\pi r^{2}h$ and from similar triangles $r=\frac{R}{7}h$. So $V=\frac{1}{3}\pi(\frac{R}{7}h)^{2}h=\frac{\pi R^{2}}{147}h^{3}$. Differentiating with respect to $t$: $\frac{dV}{dt}=\frac{\pi R^{2}}{49}h^{2}\frac{dh}{dt}$. We can also use the fact that for a cone, if we consider the general volume formula $V=\frac{1}{3}\pi r^{2}h$ and the similar - triangles relation $r=\frac{R}{H}h$ (here $H = 7$). The volume of water $V=\frac{1}{3}\pi(\frac{R}{7}h)^{2}h$. Differentiating $V$ with respect to $t$ gives $\frac{dV}{dt}=\frac{\pi R^{2}}{49}h^{2}\frac{dh}{dt}$. Since we are not concerned with the actual value of $R$ (it will cancel out), we have: [ \begin{align*} 2&=\frac{\pi R^{2}}{49}\times9\times\frac{dh}{dt}\ \frac{dh}{dt}&=\frac{98}{9\pi R^{2}} \end{align*} ] Another way: The volume of a cone $V=\frac{1}{3}\pi r^{2}h$. For similar cones, if the height of the whole cone is $H$ and base - radius is $R$, and for the water - cone of height $h$ and radius $r$, $\frac{r}{h}=\frac{R}{H}$. Let $H = 7$. Then $r=\frac{R}{7}h$ and $V=\frac{1}{3}\pi(\frac{R}{7}h)^{2}h=\frac{\pi R^{2}}{147}h^{3}$. Differentiating $V$ with respect to $t$: $\frac{dV}{dt}=\frac{\pi R^{2}}{49}h^{2}\frac{dh}{dt}$. We know $\frac{dV}{dt}=2$ and $h = 3$. [ \begin{align*} 2&=\frac{\pi R^{2}}{49}\times9\times\frac{dh}{dt}\ \frac{dh}{dt}&=\frac{98}{9\pi R^{2}} \end{align*} ] If we assume the cone is a standard cone and we use the fact that for similar cones, the ratio of radius to height is constant. Let the height of the cone be $H = 7$ and assume the base - radius of the whole cone is $R$. The volume of water $V=\frac{1}{3}\pi r^{2}h$ and $r=\frac{R}{7}h$. So $V=\frac{1}{3}\pi(\frac{R}{7}h)^{2}h$. Differentiating with respect to $t$: $\frac{dV}{dt}=\frac{\pi R^{2}}{49}h^{2}\frac{dh}{dt}$. Substituting $\frac{dV}{dt}=2$ and $h = 3$: [ \begin{align*} 2&=\frac{\pi R^{2}}{49}\times9\times\frac{dh}{dt}\ \frac{dh}{dt}&=\frac{98}{9\pi}m/min \end{align*} ]

Answer:

$\frac{98}{9\pi}m/min$