A wooden log 10 meters long is leaning against a vertical wall with its other end on the ground. The top end of the log is sliding down the

Question

A wooden log 10 meters long is leaning against a vertical wall with its other end on the ground. The top end of the log is sliding down the wall. When the top end is 6 meters from the ground, it slides down at 2m/sec. How fast is the bottom moving away from the wall at this instant?

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Tài Đức 3 years 2021-07-31T05:32:39+00:00 1 Answers 8 views 0

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  1. thongdat2
    0
    2021-07-31T05:34:33+00:00
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    Answer:

    Step-by-step explanation:

    This is a related rates problem from calculus using implicit differentiation. The main equation is Pythagorean’s Theorem. Basically, what we are looking for is \frac{dx}{dt} when y = 6 and \frac{dy}{dt}=-2.

    The equation for Pythagorean’s Theorem is

    x^2+y^2=c^2 where x and y are the legs and c is the hypotenuse. The length of the hypotenuse is 10, so when we find the derivative of this function with respect to time, and using implicit differentiation, we get:

    2x\frac{dx}{dt}+2y\frac{dy}{dt}=0 and divide everything by 2 to simplify:

    x\frac{dx}{dt}+y\frac{dy}{dt}=0. Looking at that equation, it looks like we need a value for x, y, \frac{dx}{dt} and \frac{dy}{dt}.

    Since we are looking for \frac{dx}{dt}, that can be our only unknown and everything else has to have a value. So what do we know?

    If we construct a right triangle with 10 as the hypotenuse and use 6 for y, we can solve for x (which is the only unknown we have, actually). Using Pythagorean’s Theorem to solve for x:

    x^2+6^2=10^2 and

    x^2+36=100 and

    x^2=64 so

    x = 8.

    NOW we can fill in the derivative and solve for \frac{dx}{dt}.

    Remember the derivative is

    x\frac{dx}{dt}+y\frac{dy}{dt}=0 so

    8\frac{dx}{dt}+6(-2)=0 and

    8\frac{dx}{dt}-12=0 and

    8\frac{dx}{dt}=12 so

    \frac{dx}{dt}=\frac{12}{8}=\frac{6}{4}=\frac{3}{2}=1.5 m/sec

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