Answer:
the rate that the acute angle changes at is -1/6 radians per second.
Step-by-step explanation:
Step 1:
From the diagram we can determine what y is by using the pythagoras theorem.
y² = h² - x²
y² = (15)² - (9)²
y = 12 ft
Step 2:
From the diagram we can see that the relationship that can help us determine the angle is the quotient of the opposite side and the adjacent side. Therefore,
tan(θ) = y/x
Now we must differentiate with respect to time on both sides.
(sec(θ))²(dθ/dt) = ( x(dy/dt) - y(dx/dt) ) / x²
sec(θ) = 1/ cos(θ)
= 1/(9/15)
= 5/3 rad
Step 3:
In order to determine dy/dt, we will use the pythagoras theorem.
x² + y² = h²
Now we will differentiate with respect to time.
2 x(dx/dt) + 2 y(dy/dt) = 2 h(dh/dt)
dh/dt = 0, because the length of the ladder does not change. Therefore,
2 y(dy/dt) = - 2 x(dx/dt)
dy/dt = (-2(9)(2))/ 2(12)
dy/dt = -3/2 ft/s
Step 4:
Now we have
(sec(θ))²(dθ/dt) = ( x(dy/dt) - y(dx/dt) ) / x²
(5/3)²(dθ/dt) = (9(-3/2) - 12(2)) / (9)²
dθ/dt = (9(-3/2) - 12(2)) / (9)²(5/3)²
dθ/dt = -1/6 rad/s
Therefore, the rate that the acute angle changes at is -1/6 radians per second.
Answer:
the rate that the acute angle changes at is -1/6 radians per second.
Step-by-step explanation:
From the diagram we can determine what y is by using the Pythagoras theorem.
[tex]y^{2} =h^{2} -x^{2} \\y^{2} =15^{2} -9^{2} \\y=12ft[/tex]
Step 2:
From the diagram we can see that the relationship that can help us determine the angle is the quotient of the opposite side and the adjacent side. Therefore,
tanθ [tex]= \frac{y}{x}[/tex]
Now we must differentiate with respect to time on both sides.
[tex]sec^{2}[/tex] θ = 1/cos θ
[tex]=\frac{1}{\frac{9}{15} }[/tex]
=[tex]\frac{15}{9}[/tex]
= [tex]\frac{5}{3} radian[/tex]
In order to determine [tex]\frac{dy}{dt}[/tex], we will use the Pythagoras theorem.
x² + y² = h²
Now we will differentiate with respect to time.
[tex]2x(\frac{dx}{dt}) +2y(\frac{dy}{dt} )=2h(\frac{dh}{dt} )[/tex]
[tex]\frac{dh}{dt} =0[/tex], because the length of the ladder does not change. Therefore,
[tex]2x(\frac{dx}{dt}) =-2y(\frac{dy}{dt} )[/tex]
[tex]\frac{dy}{dt} =-\frac{3}{2} ft/s[/tex]
Step 4:
Now we have
(sec(θ))²(dθ/dt) = [tex]\frac{x(\frac{dx}{dt}) -y(\frac{dy}{dt} )}{x^{2} }[/tex]
(5/3)²(dθ/dt) = (9(-3/2) - 12(2)) / (9)²
dθ/dt = [tex]\frac{9(\frac{-3}{2})-12(2) }{9^{2}(\frac{5}{3}) ^{2} }[/tex]
dθ/dt = [tex]\frac{-1}{6}[/tex] rad/s
Therefore, the rate that the acute angle changes at is -1/6 radians per
second.
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