please see screens. Please do #42 on the 2nd one.

32. A man 6 A tall is walking at the rate of 3 ft/s toward a streetlight 18 A high (see the accompanying figure).

Figure Ex-32

(a) At what rate is his shadow length changing?

(b) How fast is the tip of his shadow moving?

41-44 Find the equation of the tangent line to the graph of y = f(x) at x=x0.

41. f(x) = In x-xore-1

Answer

42. f(x) = log x: xo= 10

# Introduction:

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Calculus is a branch of mathematics that deals with the study of rates of change and the accumulation of small quantities. Calculus is essential in various fields, including physics, engineering, and economics, and it is used to solve diverse problems that involve the optimization of systems and understanding of how they change over time.

# Description:

In this mathematical exercise, we examine two problems requiring the application of calculus principles. In the first problem, a man is walking towards a streetlight, and we need to know the rate at which his shadow length is changing and the speed at which the tip of his shadow is moving. The second problem requires us to find the equation of the tangent line to the graph of the function y=f(x) at x=x0 for two given functions. By using calculus concepts such as derivatives and rates of change, we can solve these problems and understand the relationship between different variables in different contexts.

## Objectives and Learning Outcomes

### Objective 1: Related to Ex-32

– Objective: To calculate the rate of change of the man’s shadow length and tip position.

– Learning Outcomes:

– Calculate the rate of change of the man’s shadow length.

– Determine the speed of the tip of the man’s shadow.

### Objective 2: Finding the Equation of Tangent Line

– Objective: To determine the equation of tangent lines to given graphs.

– Learning Outcomes:

– Find the equation of the tangent line to a graph at a given point.

– Evaluate the slope of a tangent line to a graph.

## Heading: Ex-32

### Objective 1: Related to Ex-32

– Objective: To calculate the rate of change of the man’s shadow length and tip position.

– Learning Outcomes:

– Calculate the rate of change of the man’s shadow length.

– Determine the speed of the tip of the man’s shadow.

### Heading: 41-44 Finding the Equation of Tangent Line

### Objective 2: Finding the Equation of Tangent Line

– Objective: To determine the equation of tangent lines to given graphs.

– Learning Outcomes:

– Find the equation of the tangent line to a graph at a given point.

– Evaluate the slope of a tangent line to a graph.

#### Question 42:

– Objective: To find the equation of the tangent line to a given logarithmic graph.

– Learning Outcomes:

– Determine the value of the slope of a tangent line to a logarithmic equation.

– Find the equation of the tangent line to a logarithmic equation at a given point.

Solution 1: Man and Streetlight Problem

(a) The man’s shadow length is changing at a rate of 4.5 ft/s.

(b) The tip of his shadow is moving at a rate of 3 ft/s.

Explanation:

(a) We can use similar triangles to find the rate of change of the man’s shadow length. Let’s call the length of the man’s shadow “s”, and the distance between the man and the streetlight “d”, then we have:

s/d = (s+x)/h

where x is the length of the man, and h is the height of the streetlight. Differentiating both sides with respect to time, we get:

ds/dt * (1/d) = (dx/dt + ds/dt) * (1/h)

Rearranging and plugging in the given values, we get:

ds/dt = (4.5 ft/s)

Therefore, the man’s shadow length is changing at a rate of 4.5 ft/s.

(b) To find the rate at which the tip of his shadow is moving, we need to differentiate the equation for the length of his shadow:

s = (d*x)/h + x

Differentiating with respect to time, we get:

ds/dt = (d/dt)(x/h) + dx/dt

Plugging in the given values, we get:

ds/dt = (3 ft/s)

Therefore, the tip of his shadow is moving at a rate of 3 ft/s.

Solution 2: Equations of Tangent Lines

41. f(x) = ln(x-x0)-1

The slope of the tangent line at x=x0 is given by the derivative of the function at x=x0:

f'(x) = 1/(x-x0)

Therefore, the equation of the tangent line at x=x0 is:

y – f(x0) = f'(x0)(x – x0)

Plugging in the values, we get:

y – ln(x0-x0)-1 = 1/(x0-x0)(x-x0)

Simplifying, we get:

y = 1/(x0-x) + ln(x0-x0)-1

42. f(x) = log(x), x0 = 10

The slope of the tangent line at x=x0 is given by the derivative of the function at x=x0:

f'(x) = 1/x

Therefore, the equation of the tangent line at x=x0 is:

y – f(x0) = f'(x0)(x – x0)

Plugging in the values, we get:

y – log(10) = 1/10(x-10)

Simplifying, we get:

y = 1/10x + (log(10) – 1)

#1 Suggested Resources/Books

1. “Calculus: Early Transcendentals” by James Stewart

2. “Calculus Made Easy” by Silvanus P. Thompson

3. “Calculus: A Complete Introduction” by Hugh Neill

4. “Schaum’s Outline of Calculus” by Frank Ayres Jr.

#2 Similar Asked Questions

1. A man is walking towards a building at a certain speed. If he is 5ft tall and the building is 10ft tall, what is the rate of his shadow length change?

2. If a tree is 20ft tall and a person standing next to it is 6ft tall, how fast is the length of their shadow changing as they walk towards the tree at a rate of 3ft/s?

3. A person is standing next to a lamppost that is 12ft tall. If the person is 5ft tall and the tip of their shadow is moving at a speed of 2ft/s away from the lamppost, how fast is the person moving away from the lamppost?

4. A man is walking towards a streetlamp at a certain speed. If he is 6ft tall and the lamp is 15ft tall, what is the rate of his shadow length change?

5. A girl is walking towards a building at a certain speed. If she is 4ft tall and the building is 18ft tall, how fast is the length of her shadow changing?

#3 Find the equation of the tangent line to the graph of y = f(x) at x=x0

## 41. f(x) = In x-xore-1

The equation of the tangent line to the graph of y = ln(x – x0 – 1) at x = x0 is y = (1/(x0 – x0 – 1))(x – x0) + ln(x0 – x0 – 1).

## 42. f(x) = log x: xo= 10

The equation of the tangent line to the graph of y = log(x) at x = 10 is y = (1/10)(x – 10) + log(10).

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