4.A paper cup, which is in the shape of a right circular cone, is 16 cm deep and has a radius of 4 cm. Water is poured into the cup at a constant rate of . 3 2cm / sec


(a) At the instant the depth is 5 cm, what is the rate of change of the height?
(b) At the instant the radius is 3 cm, what is the rate of change of the radius?

Let the radius of the water level be r cm

let the height of the water be h cm
by ratio:
r/h = 4/16 = 1/4
h = 4r or r = h/4

V = (1/3)π r^2 h
= (1/3)π (h^2/16)(h) = (1/48)π r^3
dV/dt = (1/16)π r^2 dr/dt

a) for the given data:
.3 = (1/16)π(25) dr/dt
dr/dt = 16(.3)/(25π) cm/s
= ....

b) in the origininal V = ... equation, replace h to have only V and r
differentiate and sub in your values

U

(a) Well, at the instant the depth is 5 cm, we can imagine the water saying, "Hold on, let me measure my speed." So the rate of change of the height is the speed at which the water level is rising. Since the water is being poured in at a constant rate of 0.32 cm/sec, that's your answer! It's as simple as a constant stream of water, flowing smoothly and steadily.

(b) Ah, the rate of change of the radius! Like a magician pulling a rabbit out of a hat, the radius decides to change in size. At the instant the radius is 3 cm, the rate of change of the radius can be found using calculus. Specifically, you'll have to take the derivative of the radius with respect to time to determine how fast it's changing. Unfortunately, I can't do calculus, but I'm sure you'll find it fascinating!

To find the rate of change of the height and the rate of change of the radius, we can differentiate the volume of the right circular cone with respect to time.

The volume of a cone is given by the formula:
V = (1/3) * π * r^2 * h

Now let's differentiate this equation with respect to time (t):

dV/dt = (1/3) * π * (2r * dr/dt) * h + (1/3) * π * r^2 * dh/dt

We are given that dr/dt = 0 cm/s (since the radius is not changing) and dh/dt = 0.32 cm/s (constant rate of change of the depth).

Let's solve for the rate of change of the height at the instant the depth is 5 cm (h = 5 cm):

(a) At the instant the depth is 5 cm, we need to find the rate of change of the height (dh/dt) when h = 5 cm.

dV/dt = (1/3) * π * (2 * 4 * 0) * 5 + (1/3) * π * (4^2) * dh/dt

0 = 0 + (1/3) * π * 16 * dh/dt

0 = (2/3) * π * 8 * dh/dt

0 = (16/3) * π * dh/dt

Therefore, dh/dt = 0 cm/s. The rate of change of the height is 0 cm/s.

Now let's solve for the rate of change of the radius at the instant the radius is 3 cm (r = 3 cm):

(b) At the instant the radius is 3 cm, we need to find the rate of change of the radius (dr/dt) when r = 3 cm.

dV/dt = (1/3) * π * (2 * 3 * dr/dt) * 16 + (1/3) * π * (3^2) * 0.32

0 = (2/3) * π * 6 * dr/dt + (1/3) * π * 9 * 0.32

0 = (4/3) * π * dr/dt + (1/3) * π * 2.88

0 = (4/3) * π * dr/dt + (0.96/3) * π

0 = (4/3) * π * dr/dt + 0.32 * π

- 0.32 * π = (4/3) * π * dr/dt

dr/dt = - 0.32 * π / ((4/3) * π)

dr/dt = - 0.32 * 3/4

dr/dt = - 0.24 cm/s. The rate of change of the radius is -0.24 cm/s.

To solve this problem, we need to use related rates, which involves finding the relationship between different variables and their rates of change in order to find the rate of change of one variable with respect to another.

(a) To find the rate of change of the height when the depth is 5 cm, we can use the Pythagorean theorem. Here's how we can do it step by step:

Step 1: Let's call the height of the cone h and the depth of the water d.

Step 2: We know that the depth d is decreasing at a rate of 0.32 cm/sec.

Step 3: We also know that the height h, the radius r, and the depth d are related through similar triangles. The height h is proportional to the radius r, and the depth d is proportional to the radius r.

Step 4: The similar triangles give us the following relationship: h/r = d/4.

Step 5: Now, we differentiate both sides of the equation with respect to time (t) using implicit differentiation:

(dh/dt) / r - h(dr/dt) / r^2 = (dd/dt) / 4.

Step 6: We are given that dd/dt = -0.32 cm/sec (negative because the depth is decreasing), r = 4 cm, and h = 16 cm when d = 5 cm.

Step 7: Plugging in the values into the equation, we can solve for dh/dt:

dh/dt / 4 - 16(dr/dt) / 16^2 = -0.32 / 4.

dh/dt / 4 - dr/dt / 16 = -0.08.

dh/dt - dr/dt / 16 = -0.08.

dh/dt - dr/dt = -0.08 * 16.

dh/dt - dr/dt = -1.28.

Step 8: We are asked to find the rate of change of the height, so we can isolate the dh/dt term:

dh/dt = dr/dt - 1.28.

Step 9: Since we don't have the rate of change of the radius (dr/dt), we can't find the exact rate of change of the height. However, we now have an equation that relates the rate of change of the height to the rate of change of the radius.

(b) To find the rate of change of the radius when the radius is 3 cm, here's how we can do it:

Step 1: Let's call the radius of the cone r and the depth of the water d.

Step 2: We know that the radius r is decreasing at a rate of dr/dt cm/sec.

Step 3: We can use the similar triangles relationship again: h/r = d/4.

Step 4: Differentiate both sides of the equation with respect to time (t) using implicit differentiation:

(dh/dt) / r - h(dr/dt) / r^2 = (dd/dt) / 4.

Step 5: We are given that dd/dt = -0.32 cm/sec (negative because the depth is decreasing), r = 3 cm, and h = 16 cm when r = 3 cm.

Step 6: Plugging in the values into the equation, we can solve for dr/dt:

dh/dt / 3 - 16(dr/dt) / 3^2 = -0.32 / 4.

dh/dt / 3 - 16(dr/dt) / 9 = -0.08.

3(dh/dt) - 16(dr/dt) / 9 = -0.08.

Step 7: We are asked to find the rate of change of the radius, so we can isolate the dr/dt term:

-16(dr/dt) / 9 = -0.08 - 3(dh/dt).

-16(dr/dt) / 9 = -0.08 - 3(dr/dt + 1.28).

-16(dr/dt) / 9 = -0.08 - 3dr/dt - 3.84.

-16(dr/dt) / 9 + 3dr/dt = -0.08 - 3.84.

(-16 / 9 + 3/9)dr/dt = -0.08 - 3.84.

(-13 / 9)dr/dt = -3.92.

Step 8: Solve for dr/dt:

dr/dt = (-3.92) / (-13 / 9).

dr/dt = (3.92) * (9 / 13).

dr/dt ≈ 2.72/13.

dr/dt ≈ 0.209 cm/sec.

Therefore, the rate of change of the height when the depth is 5 cm is dr/dt = -1.28 cm/sec, and the rate of change of the radius when the radius is 3 cm is dr/dt ≈ 0.209 cm/sec.