Math Problem Statement

The altitude of a triangle is increasing at a rate of 1 centimeters/minute while the area of the triangle is increasing at a rate of 1.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 7 centimeters and the area is 96 square centimeters?

Solution

Let's solve this step-by-step.

Problem Breakdown:

We are given:

  1. The altitude hh of the triangle is increasing at a rate of dhdt=1cm/min\frac{dh}{dt} = 1 \, \text{cm/min}.
  2. The area AA of the triangle is increasing at a rate of dAdt=1.5cm2/min\frac{dA}{dt} = 1.5 \, \text{cm}^2/\text{min}.
  3. At the moment we are considering, the altitude h=7cmh = 7 \, \text{cm} and the area A=96cm2A = 96 \, \text{cm}^2.

We need to find: The rate at which the base bb of the triangle is changing, i.e., dbdt\frac{db}{dt}.

Key Formula:

The area of a triangle is given by: A=12bhA = \frac{1}{2} b h

Differentiate both sides with respect to time tt: dAdt=12(bdhdt+hdbdt)\frac{dA}{dt} = \frac{1}{2} \left( b \frac{dh}{dt} + h \frac{db}{dt} \right)

Solve for dbdt\frac{db}{dt}:

Rearrange the equation to isolate dbdt\frac{db}{dt}: dbdt=2hdAdtbhdhdt\frac{db}{dt} = \frac{2}{h} \frac{dA}{dt} - \frac{b}{h} \frac{dh}{dt}

Step 1: Find bb

From the area formula, A=12bhA = \frac{1}{2} b h, solve for bb: b=2Ahb = \frac{2A}{h} Substitute A=96cm2A = 96 \, \text{cm}^2 and h=7cmh = 7 \, \text{cm}: b=2(96)7=1927cm27.43cm.b = \frac{2(96)}{7} = \frac{192}{7} \, \text{cm} \approx 27.43 \, \text{cm}.

Step 2: Substitute values into the differentiated equation

Now substitute all known values into the differentiated formula: dbdt=2hdAdtbhdhdt.\frac{db}{dt} = \frac{2}{h} \frac{dA}{dt} - \frac{b}{h} \frac{dh}{dt}. Substitute h=7h = 7, dAdt=1.5\frac{dA}{dt} = 1.5, dhdt=1\frac{dh}{dt} = 1, and b27.43b \approx 27.43: dbdt=27(1.5)27.437(1).\frac{db}{dt} = \frac{2}{7}(1.5) - \frac{27.43}{7}(1).

Simplify: dbdt=3727.437.\frac{db}{dt} = \frac{3}{7} - \frac{27.43}{7}. dbdt=24.4373.49cm/min.\frac{db}{dt} = \frac{-24.43}{7} \approx -3.49 \, \text{cm/min}.

Final Answer:

The base of the triangle is decreasing at a rate of approximately: 3.49cm/min.\boxed{-3.49 \, \text{cm/min}}.

Would you like further details or have any questions?


Here are 5 related questions to expand your understanding:

  1. How does the rate of change in the area depend on both the base and height of the triangle?
  2. What happens to the base's rate of change if the altitude is constant?
  3. Can this method be extended to other geometric shapes?
  4. How would you handle this problem if the base and height were functions of time?
  5. What if the rate of change of the area was negative? How would the interpretation differ?

Tip: Always identify known rates and variables before starting the differentiation process in related rates problems.

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Math Problem Analysis

Mathematical Concepts

Related Rates
Differentiation
Geometry

Formulas

Area of a triangle: A = (1/2) * b * h
Differentiation of area with respect to time: dA/dt = (1/2) * (b * dh/dt + h * db/dt)

Theorems

Chain Rule for Differentiation

Suitable Grade Level

Grades 10-12