Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Certainly! Let's approach this problem step by step.
### Step 1: Visualize the Problem
We are given a scenario where a pedestrian is walking towards an apartment building. The height of the building is 36 meters. The pedestrian is walking at a constant speed of 60 meters per minute towards the building and is 27 meters away from the base of the building at the instant we are interested in.
### Step 2: Define the Variables
Let's label the important measurements:
- The height of the building ([tex]\( h \)[/tex]) is 36 meters.
- The distance from the pedestrian to the base of the building ([tex]\( x \)[/tex]) is 27 meters.
- The rate at which the pedestrian is walking towards the building ([tex]\( dx/dt \)[/tex]) is 60 meters per minute, but since he is approaching the building, this value will be negative: [tex]\( dx/dt = -60 \)[/tex] meters per minute.
- We need to find the rate at which the distance between the pedestrian and the top of the building ([tex]\( d \)[/tex]) is changing. Denote this rate as [tex]\( dd/dt \)[/tex].
### Step 3: Apply the Pythagorean Theorem
The distance ([tex]\( d \)[/tex]) between the pedestrian and the top of the building forms the hypotenuse of a right triangle where:
- One leg of the triangle is the height of the building ([tex]\( h \)[/tex]).
- The other leg is the horizontal distance from the pedestrian to the building ([tex]\( x \)[/tex]).
By the Pythagorean theorem:
[tex]\[ d^2 = h^2 + x^2 \][/tex]
### Step 4: Differentiate with Respect to Time
Differentiate both sides of the equation with respect to time [tex]\( t \)[/tex]:
[tex]\[ \frac{d}{dt}(d^2) = \frac{d}{dt}(h^2 + x^2) \][/tex]
[tex]\[ 2d \frac{dd}{dt} = 0 + 2x \frac{dx}{dt} \][/tex]
[tex]\[ d \frac{dd}{dt} = x \frac{dx}{dt} \][/tex]
Solving for [tex]\( \frac{dd}{dt} \)[/tex]:
[tex]\[ \frac{dd}{dt} = \frac{x \frac{dx}{dt}}{d} \][/tex]
### Step 5: Calculate the Initial Distance [tex]\( d \)[/tex]
First, we need to determine [tex]\( d \)[/tex] when [tex]\( x \)[/tex] is 27 meters:
[tex]\[ d = \sqrt{h^2 + x^2} \][/tex]
[tex]\[ d = \sqrt{36^2 + 27^2} \][/tex]
[tex]\[ d = \sqrt{1296 + 729} \][/tex]
[tex]\[ d = \sqrt{2025} \][/tex]
[tex]\[ d = 45 \text{ meters} \][/tex]
### Step 6: Plug in the Known Values
Now we have:
- [tex]\( x = 27 \text{ meters} \)[/tex]
- [tex]\( dx/dt = -60 \text{ meters per minute} \)[/tex]
- [tex]\( d = 45 \text{ meters} \)[/tex]
Substitute these into the equation:
[tex]\[ \frac{dd}{dt} = \frac{27 \times -60}{45} \][/tex]
### Step 7: Simplify the Expression
[tex]\[ \frac{dd}{dt} = \frac{-1620}{45} \][/tex]
[tex]\[ \frac{dd}{dt} = -36 \text{ meters per minute} \][/tex]
### Conclusion
The distance between the pedestrian and the top of the building is decreasing at a rate of 36 meters per minute when the pedestrian is 27 meters from the base of the building.
### Step 1: Visualize the Problem
We are given a scenario where a pedestrian is walking towards an apartment building. The height of the building is 36 meters. The pedestrian is walking at a constant speed of 60 meters per minute towards the building and is 27 meters away from the base of the building at the instant we are interested in.
### Step 2: Define the Variables
Let's label the important measurements:
- The height of the building ([tex]\( h \)[/tex]) is 36 meters.
- The distance from the pedestrian to the base of the building ([tex]\( x \)[/tex]) is 27 meters.
- The rate at which the pedestrian is walking towards the building ([tex]\( dx/dt \)[/tex]) is 60 meters per minute, but since he is approaching the building, this value will be negative: [tex]\( dx/dt = -60 \)[/tex] meters per minute.
- We need to find the rate at which the distance between the pedestrian and the top of the building ([tex]\( d \)[/tex]) is changing. Denote this rate as [tex]\( dd/dt \)[/tex].
### Step 3: Apply the Pythagorean Theorem
The distance ([tex]\( d \)[/tex]) between the pedestrian and the top of the building forms the hypotenuse of a right triangle where:
- One leg of the triangle is the height of the building ([tex]\( h \)[/tex]).
- The other leg is the horizontal distance from the pedestrian to the building ([tex]\( x \)[/tex]).
By the Pythagorean theorem:
[tex]\[ d^2 = h^2 + x^2 \][/tex]
### Step 4: Differentiate with Respect to Time
Differentiate both sides of the equation with respect to time [tex]\( t \)[/tex]:
[tex]\[ \frac{d}{dt}(d^2) = \frac{d}{dt}(h^2 + x^2) \][/tex]
[tex]\[ 2d \frac{dd}{dt} = 0 + 2x \frac{dx}{dt} \][/tex]
[tex]\[ d \frac{dd}{dt} = x \frac{dx}{dt} \][/tex]
Solving for [tex]\( \frac{dd}{dt} \)[/tex]:
[tex]\[ \frac{dd}{dt} = \frac{x \frac{dx}{dt}}{d} \][/tex]
### Step 5: Calculate the Initial Distance [tex]\( d \)[/tex]
First, we need to determine [tex]\( d \)[/tex] when [tex]\( x \)[/tex] is 27 meters:
[tex]\[ d = \sqrt{h^2 + x^2} \][/tex]
[tex]\[ d = \sqrt{36^2 + 27^2} \][/tex]
[tex]\[ d = \sqrt{1296 + 729} \][/tex]
[tex]\[ d = \sqrt{2025} \][/tex]
[tex]\[ d = 45 \text{ meters} \][/tex]
### Step 6: Plug in the Known Values
Now we have:
- [tex]\( x = 27 \text{ meters} \)[/tex]
- [tex]\( dx/dt = -60 \text{ meters per minute} \)[/tex]
- [tex]\( d = 45 \text{ meters} \)[/tex]
Substitute these into the equation:
[tex]\[ \frac{dd}{dt} = \frac{27 \times -60}{45} \][/tex]
### Step 7: Simplify the Expression
[tex]\[ \frac{dd}{dt} = \frac{-1620}{45} \][/tex]
[tex]\[ \frac{dd}{dt} = -36 \text{ meters per minute} \][/tex]
### Conclusion
The distance between the pedestrian and the top of the building is decreasing at a rate of 36 meters per minute when the pedestrian is 27 meters from the base of the building.
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
