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To solve this problem, we need to use related rates in calculus. We have a triangle whose altitude and area are changing over time. Here's the information given:

- The altitude (\( h \)) is increasing at a rate of \( \frac{dh}{dt} = 1 \text{ cm/min} \).
- The area (\( A \)) is increasing at a rate of \( \frac{dA}{dt} = 2 \text{ cm}^2/\text{min} \).
- The current altitude is \( h = 20 \text{ cm} \).
- The current area is \( A = 180 \text{ cm}^2 \).

We need to find the rate at which the base (\( b \)) of the triangle is changing, \( \frac{db}{dt} \), when the altitude is 20 cm and the area is 180 cm².

### Step 1: Area Formula
The area of a triangle is given by:
\[
A = \frac{1}{2} b h
\]
Differentiating both sides with respect to time \( t \), we get:
\[
\frac{dA}{dt} = \frac{1}{2} \left( b \frac{dh}{dt} + h \frac{db}{dt} \right)
\]

### Step 2: Plug in the Known Values
We have:
\[
\frac{dA}{dt} = 2 \text{ cm}^2/\text{min}, \quad \frac{dh}{dt} = 1 \text{ cm/min}, \quad h = 20 \text{ cm}, \quad A = 180 \text{ cm}^2
\]
We first need to find \( b \) when \( A = 180 \text{ cm}^2 \) and \( h = 20 \text{ cm} \).

Using the area formula:
\[
180 = \frac{1}{2} b (20)
\]
Solving for \( b \), we get:
\[
b = \frac{180 \times 2}{20} = 18 \text{ cm}
\]

### Step 3: Differentiate and Solve for \( \frac{db}{dt} \)
Substitute the values into the differentiated equation:
\[
2 = \frac{1}{2} \left( 18 \cdot 1 + 20 \cdot \frac{db}{dt} \right)
\]
This simplifies to:
\[
2 = \frac{1}{2} \left( 18 + 20 \frac{db}{dt} \right)
\]
Multiply both sides by 2:
\[
4 = 18 + 20 \frac{db}{dt}
\]
Subtract 18 from both sides:
\[
20 \frac{db}{dt} = -14
\]
Divide by 20:
\[
\frac{db}{dt} = -\frac{14}{20} = -0.7 \text{ cm/min}
\]

Therefore, the base of the triangle is decreasing at a rate of \( 0.7 \text{ cm/min} \).

Would you like more details or have any questions?

### Related Questions
1. How would the solution change if the altitude was decreasing instead?
2. What if the rate of change of the area was different, say 3 cm²/min?
3. Can this method be used to solve for other geometric shapes with varying dimensions?
4. What if both the altitude and base were increasing at different rates?
5. How does this problem connect to real-life applications of related rates?

### Tip
Always start related rate problems by writing down the known quantities and the equation that relates them before differentiating.

The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm²/min. At what rate is the base of the triangle changing when the altitude is 20 cm and the area is 180 cm²?

Solve a related rates problem where the altitude of a triangle increases at 1 cm/min, the area increases at 2 cm²/min, and the altitude and area are given. Learn how to calculate the rate at which the base of the triangle changes using calculus.

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