4. a spotlight is on ht is on and a 6 ft tall person is walking towards the wall at arate 0 A ft /sec. How fast is the height of the shadow on the wall Changing when the person is 8 ft from the wall? the ground 20 ft away from a wall
4. a spotlight is on ht is on and a 6 ft tall person is walking towards the wall at arate 0 A ft /sec. How fast is the height of the shadow on the wall Changing when the person is 8 ft from the wall? the ground 20 ft away from a wall
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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How can I get how fast the shadow height changes when the person is 8ft from the wall?
![**Problem Description:**
A spotlight is on the ground 20 ft away from a wall, and a 6 ft tall person is walking towards the wall at a rate of 4 ft/sec. How fast is the height of the shadow on the wall changing when the person is 8 ft from the wall?
**Diagram Explanation:**
The diagram shows a right triangle formed by the person, their shadow, and the line from the spotlight to the wall.
- Point A is where the spotlight is positioned on the ground.
- Line AB is the ground distance from the spotlight to the wall, labeled as 20 ft.
- Point D is the position of the person, and DE is the height of the person (6 ft).
- EC is the shadow on the wall, labeled as h (height of the shadow).
- x is the distance from the person to the spotlight (AD), and BD is the distance of the person from the wall.
**Mathematical Explanation:**
The relationship between the height of the person and the height of the shadow can be expressed using similar triangles:
\[
\frac{AD}{AB} = \frac{DE}{BC}
\]
Substituting the known values:
\[
\frac{x}{6} = \frac{20}{h}
\]
This implies:
\[
xh = 120
\]
Differentiating both sides with respect to time (t):
\[
\frac{dx}{dt}h + x\frac{dh}{dt} = 0
\]
Additional equations:
1. Pythagorean Theorem for any dynamic triangle considerations:
\[
a^2 + b^2 = c^2
\]
Derivatives of general power functions:
- \(\frac{d}{da}(a^n) = na^{n-1} = 2a\)
- \(\frac{d}{db}(b^n) = nb^{n-1} = 2b\)
Use these to solve the problem with the given rates and distances.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2910e309-4023-42bb-9437-090123672c12%2Fd397eca5-dd1a-4880-b12d-b1a073038c7b%2Frm9i0jb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Description:**
A spotlight is on the ground 20 ft away from a wall, and a 6 ft tall person is walking towards the wall at a rate of 4 ft/sec. How fast is the height of the shadow on the wall changing when the person is 8 ft from the wall?
**Diagram Explanation:**
The diagram shows a right triangle formed by the person, their shadow, and the line from the spotlight to the wall.
- Point A is where the spotlight is positioned on the ground.
- Line AB is the ground distance from the spotlight to the wall, labeled as 20 ft.
- Point D is the position of the person, and DE is the height of the person (6 ft).
- EC is the shadow on the wall, labeled as h (height of the shadow).
- x is the distance from the person to the spotlight (AD), and BD is the distance of the person from the wall.
**Mathematical Explanation:**
The relationship between the height of the person and the height of the shadow can be expressed using similar triangles:
\[
\frac{AD}{AB} = \frac{DE}{BC}
\]
Substituting the known values:
\[
\frac{x}{6} = \frac{20}{h}
\]
This implies:
\[
xh = 120
\]
Differentiating both sides with respect to time (t):
\[
\frac{dx}{dt}h + x\frac{dh}{dt} = 0
\]
Additional equations:
1. Pythagorean Theorem for any dynamic triangle considerations:
\[
a^2 + b^2 = c^2
\]
Derivatives of general power functions:
- \(\frac{d}{da}(a^n) = na^{n-1} = 2a\)
- \(\frac{d}{db}(b^n) = nb^{n-1} = 2b\)
Use these to solve the problem with the given rates and distances.
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