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The top of a ladder slides down a vertical wall at a rate of 0.375 m/s. At the moment when the bottom of the ladder is 6 m from the wall, it slides away from the wall at a rate of 0.5 m/s. How long is the ladder?

Respuesta :

the ladder is moving away from the wall at the bottom, namely dx/dt
at a rate of 0.5 m/s, now... when it's 6m away from the wall, it has been moving for 12seconds already then, at 0.5ms, you need 12s to get 6meters away, thus x = 6

the ladder is moving downwards, namely dy/dt
at a rate of 0.375 m/s, in 12seconds, it has moved then 4.5meters, thus y = 4.5

so hmmm, notice, the ladder is sliding down and away from the wall, but the ladder itself itself elongating or shrinking, so is a constant

let us use the pythagorean theorem for that, notice picture below
thus

[tex]\bf r^2=x^2+y^2\\\\ -----------------------------\\\\ 2r=2x\frac{dx}{dt}+2y\frac{dy}{dt}\impliedby \textit{taking common factor }2 \\\\\\ r=x\frac{dx}{dt}+y\frac{dy}{dt}\qquad \begin{cases} \frac{dy}{dt}=0.375\\\\ \frac{dx}{dt}=0.5\\\\ x=6\\ y=4.5\\\\ r=ladder \end{cases} \\\\\\ \left.\cfrac{}{} r \right|_{x=6}\implies ?[/tex]

pretty sure is very obvious :)

Ver imagen jdoe0001

The length of the ladder is [tex]\boxed{\bf 10\text{ \bf meters}}[/tex].

Further explanation:

Given:

It is given that the rate at which the ladder slides down from a vertical wall is [tex]0.375\text{ m/s}[/tex], at the same time the bottom of the ladder is [tex]6\text{ meters}[/tex] away from the wall horizontally and the horizontally sliding rate is [tex]0.5\text{ m/s}[/tex].

Calculation:

Consider the vertical length that is from the bottom of the wall to the top of the ladder as [tex]x\text{ meters}[/tex].

Consider horizontal length from the bottom of the wall to the bottom of the ladder as [tex]y\text{ meters}[/tex].

Consider the length of the ladder as [tex]Z\text{ meters}[/tex].

The top of the ladder is sliding down the wall that means the value of [tex]x[/tex] is decreasing with a rate of [tex]0.375\text{ m/s}[/tex], so the rate is as follows:

[tex]\boxed{\dfrac{dx}{dt}=-0.375}[/tex]  

The bottom of the ladder is sliding away the wall that means the value of [tex]y=6[/tex] is increasing with a rate of [tex]0.5\text{ m/s}[/tex], so the rate is as follows:

[tex]\boxed{\dfrac{dy}{dt}=0.5}[/tex]

The ladder and the wall form the right angled triangle in which [tex]x,y[/tex] is two sides the [tex]Z[/tex] is the hypotenuse of the triangle.

So according to the Pythagoras theorem, “the sum of the squares of the two sides is equal to the square of the hypotenuse of the right angled triangle.”

Now apply the Pythagoras theorem in the given problem as follows,

[tex]x^{2}+y^{2}=Z^{2}[/tex]            …… (1)

Now differentiate equation (1) to find an equation that relates rates because the information is given in rates,

[tex]2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0[/tex]  

Divide the above equation by [tex]2[/tex].

[tex]x\dfrac{dx}{dt}+y\dfrac{dy}{dt}=0[/tex]      .....(2)

Substitute [tex]-0.375[/tex] for [tex]\dfrac{dx}{dt}[/tex], [tex]0.5[/tex] for [tex]\dfrac{dy}{dt}[/tex] and [tex]6[/tex] for [tex]y[/tex] in equation (2) to obtain the value of [tex]x[/tex] as follows:

[tex]\begin{aligned}((-0.375)\times x)+(6\times 0.5)&=0\\-0.375x+3&=0\\-0.375x&=-3\\x&=\dfrac{3}{0.375}\\x&=8\end{aligned}[/tex]

So the vertical length that is from the bottom of the wall to the top of the ladder is [tex]8\text{ meters}[/tex].

Substitute [tex]8[/tex] for [tex]x[/tex] and [tex]6[/tex] for [tex]y[/tex] in equation (1) to obtain the length of the ladder as follows:

[tex]\begin{aligned}Z^{2}&=8^{2}+6^{2}\\Z^{2}&=64+36\\Z^{2}&=100\\Z&=10\end{aligned}[/tex]

 

Therefore the length of the ladder is [tex]\boxed{\bf 10\text{\bf meters}}[/tex].

Learn more:

1. Problem on rules of transformation of triangles: https://brainly.com/question/2992432

2. Problem on the triangle to show on the graph with coordinates: https://brainly.com/question/7437053

3. Problem on equivalent expression in simplified form: https://brainly.com/question/1394854

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Height and distance

Keywords: Angle, height, bottom, length, distance, triangle, 10 meters,  ladder, hypotenuse, vertical, Pythagoras theorem, sliding, sliding, 0.375m/s.