Answer :
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.
