Posted by yolana on Monday, March 14, 2011 at 7:47pm.
x = 12
x' = -13/10
y' = ? when x = 4
m = 2
y = ?
Use proportiions.
12 / 2 = 4 / y
6 = 4 / y
2/3 = y
12 / x = 4 / y
Differentiating both sides with respect to time, we get:
-12 / x^2 * x' = -4 / y^2 * y'
-12 / 4^2 * x' = -4 / (2/3)^2 * y'
-12 / 16 * x' = -9 y'
-3/4 * -13/10 = -9 y'
39 / 40 = -9 y'
-39 / 360 = y'
-0.11 m/s = y'
--------------------------------------…
y = 5
a = pi/4
a' = -pi/3
x' = ? when a = pi/4
x = ?
Let's first find x.
tan(a) = opposite / adjacent
tan(a) = y / x
tan(pi/4) = 5 / x
1 = 5/x
5 = x
Now we can rewrite as:
tan(a) = 5 / x
Now let's differentiatite both sides with respect to time. Doing so gives:
a' * sec^2(a) = -5 / x^2 * x'
-pi/3 * (1+tan^2a) = -5 / 25 * x'
-pi/3 * (1 + tan^2(pi/4)) = -1/5 * x'
-pi/3 * 2 = -1/5 * x'
-2pi/3 * -5 = x'
10pi/3 = x'
10.47 km/min = x'
Hope this helped.
sadly, it didnt work :( thanks for trying
x=h*cot(θ)
dx/dθ=-h*csc²(θ)
Given
h=5 km
dθ/dt = -π/4
θ=π/3
dx/dt=(dx/dθ)*d(θ/dt)
=-h*csc²(θ)*dθ/dt
=-5csc²(π/3)*(-π/4)
=-5*(√3)/2*(-π/4)
=1.083π/min.
=3.40 km/min (not a fast plane!)
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