#### To determine

**To find:** The rate of change of *F* with respect to θ. That is,dFdθ.

#### Answer

The rate of change of *F* with respect to θ is dFdθ=μW(sinθ−μcosθ)(μsinθ+cosθ)2_.

#### Explanation

**Given:**

The magnitute of the force is F=μWμsinθ+cosθ,

Where, μ is a costant and *W* is weight.

**Derivative Rule: Quotient Rule**

If f(θ). and g(θ) are both differentiable function, then

ddθ[f(θ)g(θ)]=g(θ)ddθ[f(θ)]−f(θ)ddθ[g(θ)][g(θ)]2 (1)

**Calculation:**

Obtain the derivative of *F.*

dFdθ=ddθ(F)=ddθ(μWμsinθ+cosθ)

Apply Quotient Rule as shown equation (1),

dFdθ=(μsinθ+cosθ)⋅ddθ[μW]−μW⋅ddθ[μsinθ+cosθ][μsinθ+cosθ]2=(μsinθ+cosθ)[0]−μW[μcosθ−sinθ][μsinθ+cosθ]2=0−μW[μcosθ−sinθ][μsinθ+cosθ]2=μW[sinθ−μcosθ][μsinθ+cosθ]2

Therefore, the rate of change of *F* with respect to θ is dFdθ=μW[sinθ−μcosθ][μsinθ+cosθ]2_.

#### To determine

**To find:** The value of θ if the rate of change equal to zero.

#### Answer

The rate of change equal to zero when θ=arctanμ_.

#### Explanation

**Given:**

From part (a), the rate of change of *F* with respect to θ is dFdθ=μW[sinθ−μcosθ][μsinθ+cosθ]2.

**Calculation:**

The rate of change equal to zero, dFdθ=0.

μW[sinθ−μcosθ][μsinθ+cosθ]2=0μW[sinθ−μcosθ]=0

Since μW≠0, sinθ−μcosθ=0.

sinθ=μcosθsinθcosθ=μtanθ=μθ=arctan(μ)

Therefore, the rate of change equal to zero when θ=arctanμ_.

#### To determine

**To draw:**. The graph of *F* as a function of θ and use it to locate the value of θ for which dFdθ=0.

#### Explanation

**Given:**

The magnitute of the force is F=μWμsinθ+cosθ, W=50 lb and μ=0.6.

Use online graphing calculator to draw the graph of *F* as a function of θ as shown in Figure 1.

The value of θ is appoximately 0.54.

**Calculation:**

Obtain the magnitute of the force *F* as a function of θ.

F=μWμsinθ+cosθ (2)

Substitute W=50 lb and μ=0.6 in equation (2),

F=(0.6)(50)(0.6)sinθ+cosθ=30(0.6)sinθ+cosθ

Therefore, the magnitute of the force *F* as a function of θ is F=30(0.6)sinθ+cosθ.

From part (b), the rate of change equal to zero when θ=arctanμ.

Substitute μ=0.6 in equation θ=arctanμ,

θ=arctan0.6≈0.54

Therefore, the value of θ is appoximately 0.54.

Substitute θ≈0.54 in s and obtain the value of *F.*

F=30(0.6)sin(0.54)+cos(0.54)=30(0.6)(0.51414)+(0.85771)=301.166194≈25.72

Therefore, the point (θ,F)≈(0.54,25.72)

Use online graphing calculator and draw the graph of *F* as a function of θ as shown below in Figure 1.

From Figure 1, it is observed that the slope of tangent is horizontanl to the curve F=30(0.6)sinθ+cosθ at the point (0.54,25.72).

Hence, the value of θ is consistent.