Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
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.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.