Mathematics, 25.06.2019 21:00, bossdde

A10-meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at 3 meters per minute. at a certain instant, the bottom of the ladder is 6 meters from the wall. what is the rate of change of the area formed by the ladder at that instant (in square meters per minute)?

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answered: tyrickdavis1

The rate of change of the area formed by the ladder at that instant is 7 square metres per minute.

Step-by-step explanation:

Let x be the length of wall. And y be the distance of ladder from the wall.

Then, the decrement in the distance between the top of the ladder and the ground is,

$\dfrac{dx}{dt}=-3$

Use Pythagoras theorem with base y and height x, and length as hypotaneous as,

$h^{2}=y^{2}+x^{2}\\10^{2}=y^{2}+x^{2}\\100=y^{2}+x^{2}$

Differentiate the above equation with respect to time as,

$100=y^{2}+x^{2}\\0=2\dfrac{dy}{dt} +2\dfrac{dx}{dt}\\-2y\dfrac{dy}{dt} =2x\dfrac{dx}{dt}\\-2y\dfrac{dy}{dt} =2x(-3)\\\dfrac{dy}{dt} =\dfrac{3x}{y}$

The area is given as,

$A=\dfrac{1}{2}xy$

Differentiate to obtain the change in area as,

$\dfrac{dA}{dt} =\dfrac{1}{2}y\dfrac{dx}{dt}+\dfrac{1}{2}x\dfrac{dy}{dt} \\\dfrac{dA}{dt} =\dfrac{1}{2}y\dfrac{dx}{dt}+\dfrac{1}{2}x\dfrac{dy}{dt}\\$

Here, y= 6 and x =8 by Pythagoras theorem. Substitute the values as,

$\dfrac{dA}{dt} =\dfrac{1}{2}y\dfrac{dx}{dt}+\dfrac{1}{2}x\dfrac{dy}{dt}\\\dfrac{dA}{dt} =\dfrac{1}{2} \times 6 \times (-3)+\dfrac{1}{2} \times 8 \times \dfrac{3x}{y}\\\dfrac{dA}{dt} =\dfrac{1}{2} \times 6 \times (-3)+\dfrac{1}{2} \times 8 \times \dfrac{3 \times 8}{6}\\\dfrac{dA}{dt} = 7$

Thus, the rate of change of the area formed by the ladder at that instant is 7 square metres per minute.

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answered: cmflores3245

Hence the rate of change in area increases by 7 square meters.

Step-by-step explanation:

Let 'y' denotes the length of the wall.

and 'x' denotes the length of the ladder from the wall.

$\dfrac{dy}{dt}$ denotes the change of the top of the ladder and the ground.

$\dfrac{dy}{dt}=-3$

now using Pythagorean theorem with base x and height y and hypotenuse is 10 ( since the length of the ladder is 10 units).

Hence,

$x^{2}+y^2=10^2\\\\x^{2}+y^2=100$

On differentiating with respect to time t we have:

$2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0\\\\2x\dfrac{dx}{dt}=-2y\dfrac{dy}{dt}\\\\2x\dfrac{dx}{dt}=-2y\times (-3)\\\\2x\dfrac{dx}{dt}=6y$

Hence, $\dfrac{dx}{dt}=\dfrac{3y}{x}$

The area is given by:

$A=\dfrac{1}{2}xy$

now the change in area is given by:

$\dfrac{dA}{dt}=\dfrac{1}{2}x\dfrac{dy}{dt}+\dfrac{1}{2}y\dfrac{dx}{dt}$

here x=6 and y=8 (by using pythagorean theorem)

we get $\dfrac{dA}{dt}=7$

Hence the rate of change in area increases by 7 square meters.

answered: Guest

x=-2

infinite number of lines, so there is an infinite number of equations.

step-by-step explanation:

there are an infinite number of line that passes through the point (-2,5)

x=-2 is one line that passes through the point

y = 5 is another line that passes through the point

(these two lines are perpendicular to each other).

since the number of numbers are infinite and the slope is the ratio of two numbers,(change in y over change in x) we can have an infinite number of slopes. our only condition is that it passes through the point (-2,5)