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A gardener's shovel is 1 m long and leaning against a fence. When the top of the shovel is sliding down the fence at a rate of 0.25 m/s. When the top of the shovel is 0.5 m off the ground, at what rate is the bottom of the shovel sliding along the ground away from the fence?

User Chaye
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Answer: To solve this problem, we can use related rates. Let's consider the following variables:

y = height of the top of the shovel above the ground (in meters)

x = distance of the bottom of the shovel from the fence along the ground (in meters)

Given:

dy/dt = rate at which the top of the shovel is sliding down the fence = 0.25 m/s

y = 0.5 m (height of the top of the shovel off the ground)

We want to find dx/dt, the rate at which the bottom of the shovel is sliding along the ground away from the fence.

Since the shovel is leaning against the fence, we can use the right triangle formed by the shovel, the fence, and the ground. Using the Pythagorean theorem:

x^2 + y^2 = 1^2

Now, differentiate both sides with respect to time (t):

2x(dx/dt) + 2y(dy/dt) = 0

Now, plug in the given values:

2x(dx/dt) + 2(0.5)(0.25) = 0

Simplify:

2x(dx/dt) + 0.25 = 0

Now, solve for dx/dt:

2x(dx/dt) = -0.25

dx/dt = -0.25 / 2x

We also know that x = √(1 - y^2), so plug in the value of y:

x = √(1 - (0.5)^2) = √(1 - 0.25) = √(0.75)

Now, calculate dx/dt:

dx/dt = -0.25 / (2 * √(0.75))

Now, simplify:

dx/dt = -0.25 / (2 * √(0.75)) ≈ -0.1443 m/s

So, the bottom of the shovel is sliding along the ground away from the fence at a rate of approximately 0.1443 m/s.

User Sansknwoledge
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