Monday, December 16, 2013

Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 38

Suppose that an object with weight w is dragged along a horizontal plane by a force acting along a rope attracted to the objet. If the rope makes an angle θ with the plane then the magnitude of the force;
F=μωμsinθ+cosθ

where μ is a constant called the coefficient of friction.
a.) Find the rate of change of F with respect to θ.
b.) When is this rate of change equal to 0?
c.) If w=50lb and μ=0.6, draw the graph of F as a function of θ and use it to locate the value of θ for which dFdθ=0. Is the value consistent with your answer to part(b)?



a.) dFdθ=[μsinθ+cosθ]ddθ(μω)μωddθ[μsinθ+cosθ][μsinθ+cosθ]2dFdθ=0μω[μcosθ+(sinθ)][μsinθ+cosθ]2dFdθ=μω[sinθμcosθ][μsinθ+cosθ]2


b.) dFdθ=0=μω[sinθμcosθ][μsinθ+cosθ]2

μω[sinθμcosθ]=0sinθμcosθ=0sinθ=μcosθsinθcosθ=μtanθ=μθ=tan1μ


The rate of change is equal to 0 when θ=tan1μ


c.) F=μωμsinθ+cosθ;ω=50h,μ=0.6F=50(0.6)0.6sinθ+cosθF=300.6sinθ+cosθ




We know that dFdθ=0 when θ=tan1μ so

θ=tan1(0.6)=0.5404F=300.6sinθ+cosθ;where θ=0.5404F(θ)=300.6sin(0.5404)+cos(0.5404)=25.725


Based from the graph, our answer in part(b) matches our answer in part(c) since the value of tangent line of that point is zero.

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