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 2.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 8 centimeters and the area is 96 square centimeters? cm/min
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 2.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 8 centimeters and the area is 96 square centimeters? cm/min
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Problem Statement:**
The altitude of a triangle is increasing at a rate of 1 centimeter per minute, while the area of the triangle is increasing at a rate of 2.5 square centimeters per minute. At what rate is the base of the triangle changing when the altitude is 8 centimeters and the area is 96 square centimeters?
---
**Explanation:**
To solve this problem, we consider the formula for the area \( A \) of a triangle:
\[ A = \frac{1}{2} \times \text{base} \times \text{altitude} \]
Given:
- The rate of change of the altitude \(\frac{dh}{dt} = 1 \) cm/min
- The rate of change of the area \(\frac{dA}{dt} = 2.5 \) cm\(^2\)/min
- The altitude \( h = 8 \) cm
- The area \( A = 96 \) cm\(^2\)
We need to find the rate of change of the base \(\frac{db}{dt}\).
**Differentiating the area formula with respect to time \( t \), we have:**
\[ \frac{dA}{dt} = \frac{1}{2} \left(b \frac{dh}{dt} + h \frac{db}{dt}\right) \]
Substitute the given values and solve for \(\frac{db}{dt}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9f1c5ee8-9f50-4f4c-aa8f-9d252d015e08%2F4a90f688-9a9e-4742-bb39-2fbbb1b43334%2Ffpovhr8_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
The altitude of a triangle is increasing at a rate of 1 centimeter per minute, while the area of the triangle is increasing at a rate of 2.5 square centimeters per minute. At what rate is the base of the triangle changing when the altitude is 8 centimeters and the area is 96 square centimeters?
---
**Explanation:**
To solve this problem, we consider the formula for the area \( A \) of a triangle:
\[ A = \frac{1}{2} \times \text{base} \times \text{altitude} \]
Given:
- The rate of change of the altitude \(\frac{dh}{dt} = 1 \) cm/min
- The rate of change of the area \(\frac{dA}{dt} = 2.5 \) cm\(^2\)/min
- The altitude \( h = 8 \) cm
- The area \( A = 96 \) cm\(^2\)
We need to find the rate of change of the base \(\frac{db}{dt}\).
**Differentiating the area formula with respect to time \( t \), we have:**
\[ \frac{dA}{dt} = \frac{1}{2} \left(b \frac{dh}{dt} + h \frac{db}{dt}\right) \]
Substitute the given values and solve for \(\frac{db}{dt}\).
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