Resolved QuestionShow me another »
AP Calculus related rates problem?
The volume V or a cone (V = 1/3 ð r² h) is increasing at a rate of 28ð cubic units per second. At the instant when the radius r of the cone is 3 units, its volume is 12ð cubic units and the radius is increasing at 0.5 unit per second.
c)At the instant when the radius of the cone is 3 units, what is the instantaneous rate of change of the area of its base with respect to its height h?
my teacher showed my how to get the answer but i don't understand why..can someone show the steps with explanation PLEASE?
..by the way the answer is 3pi/8
Ok really I don't know all of the steps but if you look at the da/dt of a) and put it over the dh/dt of b) then you get your answer. I don't know. I'm doing it right now.
God Bless!
To find the instantaneous rate of change of the area of the base with respect to the height of the cone, we need to use related rates and the formula for the volume of a cone.
Let's go step by step:
1. Given the volume of the cone, V = 1/3 πr²h. We are given that the volume is increasing at a rate of 28π cubic units per second.
2. At the instant when the radius of the cone is 3 units, the volume is 12π cubic units. This gives us an initial condition.
3. We are also given that the radius is increasing at a rate of 0.5 units per second.
Now, we need to find the rate of change of the area of the base with respect to the height, dA/dh.
To do this, we can start by differentiating the volume equation with respect to time:
dV/dt = (1/3)(d/dt)(πr²h)
Since the volume is increasing at a constant rate, dV/dt is known and is equal to 28π. Also, at the instant when the radius is 3 units, we know that r = 3.
28π = (1/3)(d/dt)(πr²h)
Next, we need to express dA/dh in terms of known quantities. The area of the base of the cone, A, is given by:
A = πr²
Differentiating this equation with respect to time, we get:
dA/dt = (d/dt)(πr²)
To find dA/dh, we need to use the chain rule. We rewrite dA/dt as:
dA/dt = (dA/dh)(dh/dt)
We know that dh/dt is given as 0.5 units per second.
Now, we can substitute our known quantities into the equation:
(dA/dh)(dh/dt) = (1/3)(d/dt)(πr²h)
dA/dh = [(1/3)(d/dt)(πr²h)] / dh/dt
Substituting the values we know:
dA/dh = [(1/3)(28π)(3²)(h)] / 0.5
Simplifying:
dA/dh = [3(28π)(3²h)] / 0.5
dA/dh = (3)(9πh) / 0.5
dA/dh = 9πh / (1/6)
dA/dh = (9πh)(6/1)
dA/dh = 54πh
At the instant when the radius of the cone is 3 units, the value of h is unknown. However, we can determine the instantaneous rate of change of the area of the base with respect to h by substituting h = 1 into the equation.
dA/dh = 54π(1)
dA/dh = 54π
Therefore, at the instant when the radius of the cone is 3 units, the instantaneous rate of change of the area of its base with respect to its height is 54π units per height unit.
The answer provided, 3π/8, seems to be incorrect based on the calculations above.