A 5 foot ladder is leaning against a vertical wall

Asked by maham237 @ 14/07/2021 in Mathematics viewed by 280 People

If the bottom of the ladder is being pulled away from the wall at the rate of 10 feet per second, at what rate is the area of the triangle formed by the wall, the ground, and the ladder changing, in square feet per second, at the instant the bottom of the ladder is 4 feet from the wall?

Answered by maham237 @ 14/07/2021

Answer:

$-11.67 ft^{2}/s$

Step-by-step explanation:

The area is given by:

$A=\frac{x.y}{2}$

The rate is given by the derivative of that function:

$\frac{dA}{dt}=\frac{1}{2} *(\frac{dx}{dt} *y + x*\frac{dy}{dt} )$ (1)

From the original question we know that

$\frac{dx}{dt}=10feet$ (2)

We also know the relation between x and y:

$y=\sqrt{5^{2}-x^{2}}$

If we derive this function and evaluate for x=4feet:

$\frac{dy}{dt} =-\frac{40}{3}$ (3)

Replacing (2) and (3) into (1):

$\frac{dA}{dt}=-\frac{70}{6}=-11.6666$

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