Answer of Calculus Home Work for School ofPharmacy

Dadang Amir Hamzah

May 7, 2013

a. Express the total loss L as a function of θ, in term of a, b, R, r and θ.Answer:From the figure we know that d1 = a − d2 cos θ and d2 = b

sin θ. So the

total loss

L = k( d1R4

+d2r4)

expressed in term of a, b, R, r and θ is

L = k(a− b cot θ

R4+b csc θ

r4).

b. Find the Critical value of θ such that the energy loss is minimal.Answer:The critical value of θ exceed when L′(θ) = 0. From answer of point a

L′(θ) = k csc2 θ( b

R4− b

r4cos θ

).

The critical value of θ is the value of θ which satisfied

k csc2 θ = 0, , cos θ =r4

R4.

So the value of θ is

θ = cos−1( r4R4

).

c. If the ratio of the pipe radii is r/R = 5/6. estimate to the nearestdegree the optimal branching angle given in part b.Answer:If we substitute the value of r = 5

6R to the answer of point b we get

the value of θ which minimized the loss function L that is

θ = cos−1( r4R4

)= cos−1

(5

6

)4 ≈ 60.49o

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d. Find the critical value of θc such that the cost of maintaining lossis minimal, which is not the same as the value θ that minimizes theprevious functional L.Answer:First lets express Lc in term of a, b, R, r and θ. Thus we get

Lc(θ) = K

[(a− b cot θ)R2 +

b

sin θr2].

By the same argument as previous problem we get

L′c(θ) = K b csc2 θ[R2 − r2 cos θ

].

The critical value of θ is the value of θ which satisfied

K b csc2 θ = 0 , R2 − r2 cos θ = 0.

So the possible value of theta is

θ = cos−1(Rr

)2.

e. We can combine these functionals, L and Lc, into one functional Lt =L+ Lc. Find the critical value θ that minimizes Lt.Answer:After we combine the previous answer we can get Lt in term of a, b, R, rand θ. The critical value of Lt exceeds when L′t(θ) = 0. By differenti-ating Lt(θ) we get

L′t(θ) = csc2 θ

[kb

R4+KR2b−

(kbr4

+ kbr2)

cos θ

].

The value of θ which minimizes Lt is the value of θ which satisfied

csc2 θ = 0 ,kb

R4+KR2b−

(kbr4

+ kbr2)

cos θ = 0.

The possible value of θ is

θ = cos−1[(

r

R

)4(kb+KbR6

kb+Kbr6

)].

f(i). By taking

limr→R

cos−1[(

r

R

)4(kb+KbR6

kb+Kbr6

)]= cos−1(1)

so we get θ = 0.

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f(ii). By taking

limR→0

cos−1[(

r

R

)4(kb+KbR6

kb+Kbr6

)]= cos−1(0)

so we get θ = π2. The two point above show that the θ agree with

original minimizer.

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