The volume of a right circular cone with radius rr and height hh is given by V=(1/3)πr^2h. The radius is increasing at a rate of 10 inches/sec, while the height is decreasing at a rate of 2 inches/sec. At what rate is the volume changing when r=12 and h=3? and go in depth on differentiation of volume please

To find the rate at which the volume of a right circular cone is changing, we will need to differentiate the volume formula with respect to time, \( t \), applying the chain rule.

The formula for the volume of a right circular cone is:

\[
V = \frac{1}{3} \pi r^2 h
\]

where:
- \( r \) is the radius,
- \( h \) is the height, and
- \( V \) is the volume.

### Step 1: Implicit differentiation of the volume formula
To find how the volume is changing over time, we differentiate both sides of the volume equation with respect to time \( t \):

\[
\frac{dV}{dt} = \frac{d}{dt} \left( \frac{1}{3} \pi r^2 h \right)
\]

We can use the product rule to differentiate the right-hand side. The product rule states that:

\[
\frac{d}{dt} (f \cdot g) = f' \cdot g + f \cdot g'
\]

Here, \( f = r^2 \) and \( g = h \), so:

\[
\frac{dV}{dt} = \frac{1}{3} \pi \left[ 2r \frac{dr}{dt} \cdot h + r^2 \cdot \frac{dh}{dt} \right]
\]

### Step 2: Substitute known values
We are given the following values:
- \( \frac{dr}{dt} = 10 \, \text{inches/sec} \) (the radius is increasing),
- \( \frac{dh}{dt} = -2 \, \text{inches/sec} \) (the height is decreasing),
- \( r = 12 \, \text{inches} \),
- \( h = 3 \, \text{inches} \).

Now, we substitute these values into the differentiated formula:

\[
\frac{dV}{dt} = \frac{1}{3} \pi \left[ 2(12)(10)(3) + (12)^2 (-2) \right]
\]

### Step 3: Simplify the expression
Let's compute each term separately.

First term:

\[
2(12)(10)(3) = 720
\]

Second term:

\[
(12)^2(-2) = 144(-2) = -288
\]

Substitute these values back:

\[
\frac{dV}{dt} = \frac{1}{3} \pi \left( 720 - 288 \right)
\]

Simplifying:

\[
\frac{dV}{dt} = \frac{1}{3} \pi (432)
\]

\[
\frac{dV}{dt} = 144 \pi \, \text{cubic inches per second}
\]

Thus, the volume is changing at a rate of \( 144\pi \, \text{cubic inches per second} \), which is approximately \( 452.39 \, \text{cubic inches per second} \).

### In-depth explanation of the differentiation process
1. **Volume formula**: We started with the volume formula \( V = \frac{1}{3} \pi r^2 h \), where the volume depends on both the radius \( r \) and height \( h \), both of which change over time.
2. **Differentiation using the product rule**: Since the volume is a product of two functions, \( r^2 \) and \( h \), we applied the product rule to differentiate the volume with respect to time.
3. **Chain rule application**: Each variable, \( r \) and \( h \), depends on time \( t \), so we used the chain rule to differentiate \( r^2 \) and \( h \), which gave us the terms involving \( \frac{dr}{dt} \) and \( \frac{dh}{dt} \).
4. **Substitution**: After differentiating, we substituted the given rates of change for the radius and height and computed the rate at which the volume is changing.

Would you like further details on any part of this differentiation process?

Here are 5 related questions to consider:
1. How would the rate of change of volume differ if both the radius and height were increasing?
2. What happens to the rate of change of the volume if the height remains constant?
3. Can you generalize this differentiation approach to a scenario where the rate of height change is a function of time?
4. How would the problem change if you were given a cone with a base that is not a circle?
5. Can you determine when the volume is decreasing based on the rates of change of radius and height?

**Tip:** Always apply the chain rule carefully when differentiating composite functions, especially when multiple variables depend on time.

The volume of a right circular cone with radius r and height h is given by V = (1/3)πr^2h. The radius is increasing at a rate of 10 inches/sec, while the height is decreasing at a rate of 2 inches/sec. At what rate is the volume changing when r = 12 and h = 3? Please go in depth on differentiation of volume.

Find out how to calculate the rate of change of the volume of a right circular cone when the radius increases by 10 inches per second and the height decreases by 2 inches per second. This solution involves implicit differentiation using the product and chain rules, and is suitable for advanced calculus students.

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