Math Problem Statement

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

Solution

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, tt, applying the chain rule.

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

V=13πr2hV = \frac{1}{3} \pi r^2 h

where:

  • rr is the radius,
  • hh is the height, and
  • VV 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 tt:

dVdt=ddt(13πr2h)\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:

ddt(fg)=fg+fg\frac{d}{dt} (f \cdot g) = f' \cdot g + f \cdot g'

Here, f=r2f = r^2 and g=hg = h, so:

dVdt=13π[2rdrdth+r2dhdt]\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:

  • drdt=10inches/sec\frac{dr}{dt} = 10 \, \text{inches/sec} (the radius is increasing),
  • dhdt=2inches/sec\frac{dh}{dt} = -2 \, \text{inches/sec} (the height is decreasing),
  • r=12inchesr = 12 \, \text{inches},
  • h=3inchesh = 3 \, \text{inches}.

Now, we substitute these values into the differentiated formula:

dVdt=13π[2(12)(10)(3)+(12)2(2)]\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)=7202(12)(10)(3) = 720

Second term:

(12)2(2)=144(2)=288(12)^2(-2) = 144(-2) = -288

Substitute these values back:

dVdt=13π(720288)\frac{dV}{dt} = \frac{1}{3} \pi \left( 720 - 288 \right)

Simplifying:

dVdt=13π(432)\frac{dV}{dt} = \frac{1}{3} \pi (432)

dVdt=144πcubic inches per second\frac{dV}{dt} = 144 \pi \, \text{cubic inches per second}

Thus, the volume is changing at a rate of 144πcubic inches per second144\pi \, \text{cubic inches per second}, which is approximately 452.39cubic inches per second452.39 \, \text{cubic inches per second}.

In-depth explanation of the differentiation process

  1. Volume formula: We started with the volume formula V=13πr2hV = \frac{1}{3} \pi r^2 h, where the volume depends on both the radius rr and height hh, both of which change over time.
  2. Differentiation using the product rule: Since the volume is a product of two functions, r2r^2 and hh, we applied the product rule to differentiate the volume with respect to time.
  3. Chain rule application: Each variable, rr and hh, depends on time tt, so we used the chain rule to differentiate r2r^2 and hh, which gave us the terms involving drdt\frac{dr}{dt} and dhdt\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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Implicit Differentiation
Related Rates

Formulas

Volume of a cone: V = (1/3)πr^2h
Product rule: d(fg)/dt = f'(t)g(t) + f(t)g'(t)
Chain rule for differentiation

Theorems

Product Rule
Chain Rule

Suitable Grade Level

College-level Calculus or Advanced High School Calculus