Asked by John
The sun is setting at the rate of 1/4 deg/min, and appears to be dropping perpendicular to the horizon, as depicted in figure 6.2.5. How fast is the shadow of a 25 meter wall lengthening at the moment when the shadow is 50 meters long?
Answers
Answered by
Damon
angle a = tan^-1 (.5)
da/dt = -.25 deg/min = -.009436 radians/min
x = 25 = y tan a
y = 25/tan a
dy/dt = -25 sec^2 a da/dt /tan^2 a
da/dt = -.25 deg/min = -.009436 radians/min
x = 25 = y tan a
y = 25/tan a
dy/dt = -25 sec^2 a da/dt /tan^2 a
Answered by
Steve
I can't see the diagram, but it appears that if the shadow is x, then
25/x = tanθ
so,
-25/x^2 dx/dt = sec^2θ dθ/dt
-25/2500 dx/dt = (1+1/4)(-1/4 * π/180)
Now just solve for dx/dt
25/x = tanθ
so,
-25/x^2 dx/dt = sec^2θ dθ/dt
-25/2500 dx/dt = (1+1/4)(-1/4 * π/180)
Now just solve for dx/dt
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