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A 10​-ft ladder is leaning against a house when its base starts to slide away. by the time the base is 6 ft from the​ house, the base is moving away at the rate of 8 ​ft/sec.
a. what is the rate of change of the height of the top of the​ ladder?
b. at what rate is the area of the triangle formed by the​ ladder, wall, and ground changing​ then? a coordinate plane has a horizontal x-axis and a vertical y-axis. a thin rectangle falling from left to right labeled 10 foot ladder starts on the y-axis at y(t) and ends on the x-axis at x(t). an arrow on the y-axis below y(t) is pointing down and an arrow on the x-axis to the right of x(t) is pointing to the right. the angle created between the segment and the x-axis is labeled theta. theta x(t) y(t) 0 1

Respuesta :

CastleRook
a. what is the rate of change of the height of the top of the​ ladder?
Let y be the height
Let x be the base
dx/dt=8 ft/sec, x=6 ft, hypotenuse=10 ft
x²+y²=10²
2x(dx/dt)+2y(dy/dt)=0
solving for dy/dt in terms of x,y and dx/dt:
dy/dt=(-x/y)(dx/dt)
but now x=6 and y=8
dy/dt=(-6/8)(8)=-6 ft/sec

b]b. at what rate is the area of the triangle formed by the​ ladder, wall, and ground changing​ then?
Here the rate of change of the area is dA/dt

dA/dt=1/2(x*dy/dt+y*dx/dt)
but
x=6, y=8, dy/dt=-6, dx/dt=8
plugging in our values we get:
dA/dt=1/2(6×(-6)+8(8))
dA/dt=1/2(-36+64)=14 ft²/sec

c. At what rate is the angle between the ladder and the ground changing then?
The relationship that relates the angle with sides x and y of a right angle:

dθ/dt=-1/sinθ1/10dx/dt
sinθ=8/10
dx/dt=8
thus
-10/8×1/10×8
=-1 rad/sec

a) [tex]\rm \dfrac{dy}{dt} = -6 ft/sec[/tex]

b) [tex]\rm \dfrac{dA}{dt}= 14ft^2/sec[/tex]

c) [tex]\rm \dfrac{d\theta}{dt}= -1\; rad/sec[/tex]

Step-by-step explanation:

Given :

Ladder height = 10ft

Base is moving away at the rate of 8 ​ft/sec.

Calculation :

Let y be the height  and x be the base.

[tex]\rm \dfrac{dx}{dt}= 8ft /sec[/tex]

x = 6ft

Applying pythagorean theorem

[tex]x^2 +y^2= 10^2[/tex]  ------ (1)

Differentiate equation (1) with respect to time t,

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

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

a) Now in equatiuon (2) putting the value of x = 6 and y = 8 we get ,

[tex]\dfrac{dy}{dt}=-\dfrac{6}{8}\times 8[/tex]

[tex]\rm \dfrac{dy}{dt}= -6ft/sec[/tex]

b) The rate of change of the area is

[tex]\rm \dfrac{dA}{dt}[/tex]

[tex]\rm \dfrac{dA}{dt}= \dfrac{1}{2}(x \times \dfrac{dy}{dt}+y\times\dfrac{dx }{dt})[/tex] ----- (3)

Now in equation (3) putting the value of

[tex]x = 6 ,\; y=8,\;\dfrac{dx}{dt}=8,\;\dfrac{dy }{dt}=-6[/tex]

we get,

[tex]\rm \dfrac{dA}{dt}=\dfrac{1}{2}(6\times(-6)+8\times8)[/tex]

[tex]\rm \dfrac{dA}{dt}=14ft^2/sec[/tex]

c) Relationship that relates the angle with sides x and y of a right angle,

[tex]\rm \dfrac{d\theta}{dt}=-\dfrac{1}{sin\theta}\times\dfrac{1}{10}\times \dfrac{dx}{dt}[/tex]  ------- (4)

Now in equation (4) putting the value of

[tex]\rm sin\theta = \dfrac{8}{10}, \; \dfrac{dx}{dt}=8[/tex]

we get,

[tex]\rm \dfrac{d\theta}{dt}=-\dfrac{10}{8}\times\dfrac{1}{10}\times 8[/tex]

[tex]\rm \dfrac{d\theta}{dt}= -1 \;rad/sec[/tex]

For more information, refer the link given below

https://brainly.com/question/24898810?referrer=searchResults

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