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APPLICATION OF DERIVATIVES

Question 1. A window is being built and the bottom is a

rectangle and the top is a semicircle. If there is 12 meters offraming materials what must the dimensions of the window

be to let in the most light?

(A)3.3606 x 1.6803

(B)3.3606 x 0.8401

(C)1.6803 x 1.6803

(D)0.8401x 0.8401

Answer: A

Solution:

The height of the rectangular portion is h and because the

semicircle is on top we can think of the width of the

rectangular portion at 2r.

The perimeter (our constraint) is the lengths of the threesides on the rectangular portion plus half the circumference

of a circle of radius r. The area (what we want to maximize)

is the area of the rectangle plus half the area of a circle of

radius r.

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Solution: Conclusion of the Mean Value Theorem.

f(15)-f(6)=f'(c)(15-6)

Plugging in for the known quantities and rewriting this a little

gives,

f(15)=-2+9f'(c)

f'(x)

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L(x)=f(a)+f'(a)(x-a)

f'(x)=(1/3)x^-2/3

f'(x)=1/(3x^2/3)

f(8)=2

f'(8)=1/12

L(x)=2+[1/12(x-8)]=x/12+4/3

L(8.05)=2.00416

L(25)=3.4166

Sum=5.42076

LOD: Medium

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Question 4. A sphere was measured and its radius was

found to be 45 cm with a possible error of no more that 0.01

cm. What is the maximum possible error in the volume if weuse this value of the radius?

(A)250 cm^3

(B)254 cm^3

(C) 255 cm^3

(D) 257 cm^3

Answer: C

Solution: Volume of a sphere: 4/3(pi*r^3)

dV=4*pi*r^2 dR

delta V=dV

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Given R=45

dv=4*pi*(45)^2* 0.01= 255 cm^3

Hence the error in measuring the volume is ~ 255 cm ^3

LOD: Easy

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Question 5. Sand is pouring from the pipe at the rate of 12

cc/sec.The falling sand forms a cone on the ground in such a

way that height of the cone is always 1/6th of the radius of

the base of the cone.How fast is the height of sand coneincreasing when the height is 4 cm.

(A)2.14 cm/sec

(B)0.0072 cm/sec

(C)0.036 cm/sec

(D) 3.14 cm/sec

Answer: B

Solution: Volume of cone=V= 1/3*pi*R^2*H

Let 1/3*pi=k

(dV/dt)= 2kRH(dR/dt)+kR^2(dH/dt).......................(eq.1)

(dV/dt)= 12 cc/sec

H=R/6at any given time 't'

(dR/dt)=6*(dH/dt)

when H=4 ,R is 24

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putting the values in the eq 1.

12=2k*24*4*6*(dH/dt)+k*576*(dH/dt)

(dH/dt)=12/(1728k)

k=22/21

(dH/dt)=0.0072 cm/sec

LOD: Medium

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