Mathematics, 28.08.2019 03:30 organicmemez

Aladder 25 feet long is leaning against
the wall of a house. the base of the ladder is
pulled away from the wall at a rate of 2 feet per second
(a) how fast is the top of the ladder moving down the wall
when its base is 7 feet, 15 feet, and 24 feet from the wall?
(b) consider the triangle formed by the side of the house, the
ladder, and the ground. find the rate at which the area
of the triangle is changing when the base of the ladder is
7 feet from the wall.
(c) find the rate at which the angle between the ladder and the
wall of the house is changing when the base of the ladder
is 7 feet from the wall.

Answers: 1

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Try to reproduce this problem in your head. Imagine a ladder leaning against a wall of your house. Notice that it forms a right triangle, where the ladder is the hypothenuse.

Basically, the relation of that right triangle is

$x^{2} +y^{2} =25^{2}$ , which is Pythagorean Theorem.

It's important to notice that this problem is about "how fast" does something changes, which refers to derivatives. For example, if you read question (a), you'll notice that is asking for the velocity of change of the top of the ladder. All this means that we need to calculate derivative, also and are functions of the time: x(t) and y(t) , because they change through time.

So, the first expression must change to: $x(t)^{2} + y(t)^{2} =25^{2}$

If we calculate the derivative of this expression would be:

$\frac{d}{dt} [x(t)^{2} + y(t)^{2}] =\frac{d}{dt} [25^{2}]=0$ , because the derivative of a constant is zero.

The other derivatives are solved using the change rule, because they are combined functions:

$\frac{d}{dt}[x(t)^{2}]=2x(t) \times x'(t)=2xx' \\\frac{d}{dt}[y(t)^{2}]=2y(t) \times y'(t)=2yy'$

So, the initial equation with the derivatives, is

2xx'+2yy'=0

We need to find the change regarding the vertical variable , so

$2xx'+2yy'=0\\2yy'=-2xx'\\y'=\frac{-2xx'}{2y} \\y'=-\frac{xx'}{y}$

Now, we know that distance is gonna be 7 feet, 5 feet and 24 feet. Let's find the respective distance to each of them, by using the first equation:

If x=7 , then $x^{2} +y^{2} =25^{2} \implies y^{2} =25^{2} - 7^{2}=576 \implies y=24$

If x=15 , then $x^{2} +y^{2} =25^{2} \implies y^{2} =25^{2} - 15^{2}=400 \implies y=20$

If x=24 , then $x^{2} +y^{2} =25^{2} \implies y^{2} =25^{2} - 24^{2}=49 \implies y=7$

All these values are needed to evalutate the expression:

$y'=-\frac{xx'}{y}$ , to find the change in each case.

Remeber that the rate is 2 feet per second, that means x'=2 .

Replacing all values in all three cases, we have

$y'=-\frac{xx'}{y}=-\frac{7(2)}{24} =-\frac{14}{24} =-\frac{7}{12} \\y'=-\frac{xx'}{y}=-\frac{15(2)}{20} =-\frac{30}{20} =-\frac{3}{2}\\y'=-\frac{xx'}{y}=-\frac{24(2)}{7} =-\frac{48}{7}$

Therefore,

When its base is 7 feet, the top of the ladder moves down the wall 7/12 feet per second.When its base is 15 feet, the top of the ladder moves down 3/2 feet per second.When its base is 24 feet, the top of the ladder moves down 48/7 feet per second.

Notice that part (b) is about the area of the triangle, which is defined as

$A(t)=\frac{1}{2}x(t) y(t)$

Where x(t) represents the base (the horizontal distance), and y(t) represents height (vertical distance). Also, remember that both variables change through time, that's why they need to depend on time.

Then, we calculate the derivative of this expression

$\frac{d}{dt}[A(t)]=\frac{d}{dt}[\frac{1}{2}x(t) y(t)]$

Here we need to use the product property for derivative, which states

$\frac{d}{dt}[f(t)g(t)] =f'(t)g(t)+f(t)g'(t)$

Using the property, we have

$A'(t)=\frac{1}{2}[x'(t)y(t)+x(t)y'(t)]$

But, we know that the ladder moves at a rate of 2 feet per second, and we know the value of the other variables, replacing them, we have

$A'(t)=\frac{1}{2}[2(24)+(7)(-\frac{7}{12} )] =\frac{1}{2}[48-\frac{48}{12} ]=\frac{527}{24}$

Therefore, when the base is 7 feet long, the rate at which the area of the triangle is changing is 527/24 square feet per second. (b)

For (c), the angle between the ladder and the wall is gonna be $\theta$ , if we use the sine function, the relation would be

$sin(\theta)=\frac{Opposite \ Leg}{Hypothenuse}=\frac{x}{25}$

But, the rate of the angle changing indicates we need to find its derivative:

$\frac{d}{dt}[sin(\theta(t))]=\frac{d}{dt}[\frac{x(t)}{25} ]$ , because functions depend on time to change.

So, the derivative is $cos(\theta) \times \theta' = \frac{1}{25} x'$ , solving for $\theta'$ we have

$\theta'=\frac{x'}{25cos\theta}$

Where cosine function is defined as: $cos\theta = \frac{y}{25}$

So, replacing that, we have

$\theta'=\frac{x'}{25\frac{y}{25} } =\frac{x'}{y}$

But, we know that the rate is 2 feets per second, that is x'=2 and y=24 .

Then,

$\theta'=\frac{2}{24}=\frac{1}{12}$

Therefore, the angle between the ladder and the wall changes at a rate of 1/12 radians per second.

Answer from: Quest

answer: y=4000x+14000

step-by-step explanation:

Estimate an equation for the line of best fit for the following scatterplot

Answer from: Quest

answer

d. 10

explanation

the distance from point y to wx is the height of the triangle.

using pythagoras theorem,

$(zy )^{2} +10^{2} = {(10 \sqrt{2}) }^{2}$

$(zy )^{2} +100= 200$

group similar terms to get,

$(zy )^{2}= 200 - 100$

this implies that,

$(zy )^{2}=100$

take positive square root of both sides to get,

$zy = \sqrt{100}$

this implies that,

$zy = 10$

What is the distance from point y to wx in the figure below

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Aladder 25 feet long is leaning against
the wall of a house. the base of the ladder is
p...