A flower pot is dropped from a point 12 feet above the ground,so that it will land 4 feet from the base of a 12 foot lamppost. At what rate is the shodow of the flower pot moving along the ground when the pot is 8 feet above the ground?
height y = 12 - 16t^2
shadow at x, similar triangles, so
x/12 = (x-4)/y
or,
xy = 12x - 48
y dx/dt + x dy/dt = 12 dx/dt
y=8 when t=1/2
x=6 when y=8
(12 - 16t^2) dx/dt + x (-32t) = 12 dx/dt
at t=1/2,
8 dx/dt - 96 = 12 dx/dt
dx/dt = 96/-4 = -24 ft/s
Oops.
x=12 when y=8
make the change and percolate it through the rest of the solution.
To solve this problem, we can use similar triangles and the concept of related rates.
Let's consider the following diagram:
```
F (flower pot)
/|
/ |
/ |x (shadow of the flower pot)
/ |
/ |
/_____|______
L (base of the lamppost)
```
Note: The symbol 'x' represents the length of the shadow of the flower pot.
We have two similar triangles: FLT (top triangle) and LGS (bottom triangle).
From the problem's description, we know the following measurements:
- Height LT = 12 feet
- Distance FL = 4 feet
- Height FG (flower pot above the ground) = 8 feet
Using similar triangles, we can set up the following proportions:
LT / FL = LG / GS
Substituting the given values:
12 / 4 = HG / x
Now we can solve for x:
12x = 4HG
x = 4HG / 12
x = HG / 3
To find the rate at which the shadow is moving along the ground, we need to differentiate the equation x = HG / 3 with respect to time. This will give us the rate of change of x with respect to time.
Differentiating the equation x = HG / 3 implicitly with respect to time will give:
(dx/dt) = (dHG/dt) / 3
Now we need to find dHG/dt, which represents the rate at which the height of the flower pot is changing with respect to time.
Given that the flower pot is dropping, its height with respect to time can be described by the equation:
HG^2 = FG^2 + FL^2
Differentiating both sides of the equation implicitly with respect to time will give:
2HG(dHG/dt) = 2FG(dFG/dt)
Since we want to find dHG/dt, we can solve for it by substituting the values we know into the equation:
2HG(dHG/dt) = 2FG(dFG/dt)
2HG(dHG/dt) = 2(8)(dFG/dt)
2HG(dHG/dt) = 16(dFG/dt)
Rearranging the equation:
(dHG/dt) = 8(dFG/dt) / HG
Now we have all the pieces to calculate the rate at which the shadow is moving along the ground when the flower pot is 8 feet above the ground.
1. Calculate (dHG/dt) using the equation:
(dHG/dt) = 8(dFG/dt) / HG
2. Substitute the value of HG = 8 feet and solve for (dHG/dt).
3. Substitute the calculated value of (dHG/dt) and HG = 8 into the equation:
(dx/dt) = (dHG/dt) / 3
4. Calculate the value of (dx/dt) to find the rate at which the shadow is moving along the ground when the flower pot is 8 feet above the ground.