Question

A 25-meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at 3.5 meters per minute. At a certain instant, the top of the ladder is 7 meters from the ground. What is the rate of change of the distance between the top of the ladder and the ground at that instant (in meters per minute)?

Answer 1

Answer:

Step-by-step explanation:

This situation models a right triangle.  The hypotenuse is the ladder which has a given length of 25 meters.  Let's call the ground (the base of the triangle) x, and the vertical wall (the height of the triangle) y.  Pythagorean's Theorem tells us that

$x^2+y^2=25^2$

This is our main formula.

We are given that the rate the distance between the bottom of the ladder and the wall is increasing at 3.5 m/min.  This is the rate of change of x, or dx/dt.  We are wanting to find the rate of change of the distance between the top of the ladder and the ground, dy/dt, when y = 7.  Let's find the derivative of our equation first, then fill in what we are given:

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$

From our given, we can fill in the dx/dt, and the y.  We are looking for dy/dt, but we are missing what the x value is.  Let's find that by going back to Pythagoreans Theorem and filling in y to find x:

$x^2+7^2=25^2$ and

$x^2=576$ so

x = 24

Now let's fill in our derivative:

$2(24)(3.5)+2(7)\frac{dy}{dt}=0$ and

$168+14\frac{dy}{dt}=0$ and

$14\frac{dy}{dt}=-168$ so

$\frac{dy}{dt}=-12m/min$

Within the context of our problem, this means that the height of the ladder is decreasing at a rate of 12 m/min.  The negative value lets us know that the height is decreasing, not that the height is negative.

Answer 2

This question is based on the Pythagorean's Theorem. Therefore, the rate of change of the distance between the top of the ladder and the ground at that instant  is -12 m/min.

Given:

25-meter ladder is sliding and the wall is increasing at 3.5 meters per minute. At a certain instant, the top of the ladder is 7 meters from the ground.

We have to calculate the rate of change of the distance between the top of the ladder and the ground at that instant (in meters per minute).

According to the question,

Diagram is attached below.

It forms a right triangle.  The hypotenuse is the ladder which has a given length of 25 meters and the ground (the base of the triangle) x, and the vertical wall (the height of the triangle) y.  

By Pythagorean's Theorem,

$x^{2} +y^{2} =25^2$

$x^{2} + 7^2 = 25^2\\x^2 + 49 =625\\x^2 = 625 - 49\\x^2 = 576\\x = 24$

Therefore, the base x is equal to 24.

Let's find the derivative of above equation.

We get,

$2x \dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$

It is given that,  the wall is increasing at 3.5 meters per minute that means height i.e. y is increasing.

Now put all the value in equation$2x \dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$.

We get,

$2(24)(3.5)+2(7)(\dfrac{dy}{dt} ) = 0\\\\168+ (14)(\dfrac{dy}{dt} )=0\\\\\dfrac{dy}{dt} = \dfrac{-168}{14} \\\\\dfrac{dy}{dt} = -12 m/min$

Therefore, the rate of change of the distance between the top of the ladder and the ground at that instant  is -12 m/min.

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