Question

at noon, ship a is 100 km west of ship b. ship a is sailing south at 40 km/h and ship b is sailing north at 20 km/h. how fast (in km/h) is the distance between the ships changing at 4:00 p.m.? (round your answer to one decimal place.)

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water is leaking out of an inverted conical tank at a rate of 10,500 cm³/min at the same time that water is being pumped into the tank at a constant rate. the tank has height 6 m and the diameter at the top is 4 m. if the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate (in cm³/min) at which water is being pumped into the tank. (round your answer to the nearest integer.)

Explanation:

Step1: Set up variables for ship problem

Let $x$ be the distance ship A has sailed south, $y$ be the distance ship B has sailed north. Initially, the horizontal distance between them is $100$ km. At time $t$ (in hours), $x = 40t$, $y=20t$. The distance $z$ between the two ships is given by the Pythagorean - theorem $z^{2}=(100)^{2}+(x + y)^{2}$.

Step2: Differentiate the equation with respect to time $t$

Differentiating $z^{2}=10000+(x + y)^{2}$ with respect to $t$ using the chain - rule, we get $2z\frac{dz}{dt}=2(x + y)(\frac{dx}{dt}+\frac{dy}{dt})$.

Step3: Calculate $x$, $y$, and $z$ at $t = 4$

At $t = 4$ hours, $x=40\times4 = 160$ km, $y = 20\times4=80$ km. Then $x + y=160 + 80=240$ km, and $z=\sqrt{100^{2}+240^{2}}=\sqrt{10000 + 57600}=\sqrt{67600}=260$ km. Also, $\frac{dx}{dt}=40$ km/h and $\frac{dy}{dt}=20$ km/h.

Step4: Solve for $\frac{dz}{dt}$

Substitute $z = 260$, $x + y=240$, $\frac{dx}{dt}=40$, and $\frac{dy}{dt}=20$ into $2z\frac{dz}{dt}=2(x + y)(\frac{dx}{dt}+\frac{dy}{dt})$. We have $2\times260\times\frac{dz}{dt}=2\times240\times(40 + 20)$. Then $\frac{dz}{dt}=\frac{240\times60}{260}\approx55.4$ km/h.

Step5: Set up variables for tank problem

Let $V$ be the volume of water in the tank, $h$ be the height of the water in the tank. The ratio of the radius $r$ to the height $h$ of the large cone is $\frac{r}{h}=\frac{2}{6}=\frac{1}{3}$ (since the diameter of the large cone at the top is $4$ m and height is $6$ m), so $r=\frac{1}{3}h$. The volume of a cone is $V=\frac{1}{3}\pi r^{2}h$. Substituting $r=\frac{1}{3}h$ into the volume formula, we get $V=\frac{1}{3}\pi(\frac{1}{3}h)^{2}h=\frac{1}{27}\pi h^{3}$.

Step6: Differentiate the volume formula with respect to time $t$

Differentiating $V=\frac{1}{27}\pi h^{3}$ with respect to $t$ using the chain - rule, we have $\frac{dV}{dt}=\frac{1}{9}\pi h^{2}\frac{dh}{dt}$.

Step7: Calculate $\frac{dV}{dt}$ when $h = 200$ cm and $\frac{dh}{dt}=20$ cm/min

When $h = 200$ cm, $\frac{dV}{dt}=\frac{1}{9}\pi(200)^{2}\times20=\frac{800000\pi}{9}$ $cm^{3}/min\approx279252.7$ $cm^{3}/min$. Let the rate at which water is pumped in be $P$ and the rate at which water is leaking out be $L = 10500$ $cm^{3}/min$. We know that $\frac{dV}{dt}=P - L$.

Step8: Solve for $P$

$P=\frac{dV}{dt}+L$. Substituting $\frac{dV}{dt}=\frac{800000\pi}{9}$ and $L = 10500$, we get $P=\frac{800000\pi}{9}+10500\approx279252.7+10500=289753$ $cm^{3}/min$.

Answer:

• For ship problem: $55.4$ km/h
• For tank problem: $289753$ $cm^{3}/min$