Asked by Rachel
If h= 65 tan theta, what is the approximate change in h when theta changes from 60° to 60° 03'?
Answers
Answered by
oobleck
since dy/dx ≈ ∆y/∆x
∆y ≈ dy/dx ∆x
so for your problem, since 3' = 1/20 = 0.05°
∆θ ≈ 65 sec^2(60°) * 0.05 = 65*4 * 0.05 = 13
that makes 65tan(60° 03') ≈ 65√3 + 13 = 125.58
check: 65tan(60° 03') = 112.81
not very close, but the graph is curving pretty sharpy at this point, making the linear approximation not especially good.
∆y ≈ dy/dx ∆x
so for your problem, since 3' = 1/20 = 0.05°
∆θ ≈ 65 sec^2(60°) * 0.05 = 65*4 * 0.05 = 13
that makes 65tan(60° 03') ≈ 65√3 + 13 = 125.58
check: 65tan(60° 03') = 112.81
not very close, but the graph is curving pretty sharpy at this point, making the linear approximation not especially good.
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