Question

A 10-foot ladder is leaning straight up against a wall when a person begins pulling the base of the ladder away from the wall at the rate of 1 foot per second. Calculate the true distance between the top of the ladder and the ground when the base of the ladder is 9 feet from the wall.

Answer 1

Answer:

$y = 4.36$

Explanation:

Let the height of the ladder be L

$L = 10$

Also:

• Let $x = distance\ from\ the\ base\ of\ the\ ladder$

• Let $y = height\ of\ the\ base\ of\ the\ ladder$

When the ladder leans against the wall, it forms a triangle and the length of the ladder forms the hypotenuse.

So, we have:

$L^2 = x^2 + y^2$ --- Pythagoras Theorem

When the base is 9ft from the wall, this means that:

$x = 9$

Substitute 9 for x and 10 for L in $L^2 = x^2 + y^2$

$10^2 = 9^2 + y^2$

$100 = 81 + y^2$

Make $y^2$ the subject

$y^2 = 100 - 81$

$y^2 = 19$

Make y the subject

$y = \sqrt{19$

$y = 4.36$

Hence, the true distance at that point is approximately 4.36ft

Answer 2

The true statement about the true distance between the top of the ladder and the ground when the base is 9 feets from the wall is - 9/√19

Let :

• x = distance from base of ladder
• y = Length of ladder

Ladder forms a right angle triangle as it leans against the wall :

Hence, we have :

• Opposite² + Adjacent² = hypotenus²

• x² + y² = 100

Taking the implicit derivative of both sides:

2xdx/dt + 2ydy/dt = 0

xdx/dt + ydy/dt = 0

ydy/dt = - xdx/dt

dy/dt = (-xdx/dt) / y

When the ladder is 9 ft from the wall;

• x = 9

81 + y² = 100

y² = 100 - 81 = 19

y = √19

Substituting into the equation :

dy/dt = -9(1)/√19

Hence, dy/dt = - 9/√19

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