Consider that there is a container having the shape of an inverted cone. It has a base- diameter of 24cm and the height is twice as its diameter. The inverted cone contains the liquid and due to a small leakage at the bottom end, liquid flows out at the rate of 8ml of every second. Find the rate of change of the height of the liquid when the height of liquid is 19cm above its bottom end.
can you elaborate please? i dont get how you can find the final answer
really? I showed every step.
Read up on implicit differentiation, and look in your text, or online, for "related rates" examples.
sorry about the typo. That V should have been π (did shift-v instead of ctrl-v)
To find the rate of change of the height of the liquid, we can use the concept of similar triangles and rate of change of volume.
Let's start by finding the volume of the inverted cone. The formula for the volume of a cone is V = (1/3) * π * r^2 * h, where r is the radius of the base and h is the height.
Given that the base diameter is 24cm, we can find the radius by dividing the diameter by 2:
r = 24 / 2 = 12cm
The height is twice the diameter:
h = 2 * 24 = 48cm
Substituting these values into the volume formula, we get:
V = (1/3) * π * 12^2 * 48
V = 2π * 12^2 * 48
V = 1152π cm^3
Now, let's consider the rate of change of volume with respect to the height. We want to find the rate of change of the height, which is dh/dt, when the height is 19cm. We want to find this rate when the volume is decreasing at a rate of 8ml/s.
Since we have the volume V and we know the rate of change of volume dV/dt, we can find the rate of change of height dh/dt by differentiating the volume formula with respect to time.
dV/dt = (d/dt)[(1/3) * π * r^2 * h]
Now, differentiating the formula, we get:
dV/dt = (1/3) * π * [2 * r * (dr/dt) * h + r^2 * (dh/dt)]
Since the base diameter is given, we can find the rate of change of the radius (dr/dt) based on the given rate of change of volume.
We know that the volume is decreasing at a rate of 8ml/s, which is equivalent to 8 cm^3/s because 1ml = 1cm^3. Hence, dV/dt = -8 cm^3/s.
Substituting this value into the equation, we get:
-8 = (1/3) * π * [2 * 12 * (dr/dt) * 48 + 12^2 * (dh/dt)]
Simplifying further, we have:
-8 = 48π * (dr/dt) + 576π * (dh/dt)
Now, we know that when the height of the liquid is 19cm, we want to find the rate of change of the height (dh/dt). So let's substitute the known values into the equation:
-8 = 48π * (dr/dt) + 576π * (dh/dt)
dh/dt = (-8 - 48π * (dr/dt)) / 576π
To find the value of dr/dt, we need more information about the leak or how the volume is changing. Without additional information, we cannot determine the rate of change of the height of the liquid when it is 19cm above the bottom end.
using similar triangles, when the depth of the liquid is h, then the radius of the surface is h/4
Thus, the volume of the liquid is
v = 1/3 V(h/4)^2 h = π/48 h^3
dv/dt = π/16 h^2 dh/dt
when h=19 cm,
-8 = π/16 * 19^2 dh/dt
dh/dt = -128/(361π) cm/s