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Civil Engineering Board Exam

(List of Subjects)

2006une,ath surveying and transpo

-limits- , &area centroid volume of the formed area of the given curves- 'ellipse properties- ( . ****- ,inverse matrix eto pa ang hindi ko nasagutan hehe ina hindi alam gamitin calcu sa

.inverse matrix )- /dy dx of given formulas- 'circle s properties-simple curve-straight line depreciation-projectile motion

/ydraulics geotechnical-water hammer-lateral earth pressure- -tri axial test- ( !) ,BUOYANCY so damn bonus next question compute for specific weight of the metal -relative density-properties of soil

-flow of water through soil-headloss on pipes- ( )orifice time to discharge a tank-pile capacity

...medyo humirap lang sa designdesign- ( )torsional moment of the beam i think mali yung given formula- ( , ' 'concrete mix wala sa choices ang answer i don t know if it s bonus o ako lang ang

. )mali hehe-cantilever method- , , .shear moment diagram compute reactionmax moment -angular deformation-brackets-basic construction estimate

- ,adequacy of beam in shearing bending stress and deflection-two equivalent set of concurrent forces

references:

• http://www.analyzemath.com/calculus/Problems/rate_change.html

Use the Chain Rule of Differentiation in Calculus

Find the derivative f '(x), if f is given by



Example 1: f(x) = 4 cos (5x - 2)Solution to Example 1

• Let u = 5x - 2 and y = 4 cos u, hence du/dx = 5 and dy/du = -4 sin u

• We now use the chain rule f '(x) = (dy/du) (du/dx) = - 4 sin (u) (5)

• We now substitute u = 5x - 2 in sin (u) above to obtain

f '(x) = - 20 sin (5x – 2)

Example 2: f(x) = (x 3 - 4x + 5) 4 Solution to Example 2

• Let u = x 3 - 4x + 5 and y = u 4 , hence

du/dx = 3 x 2 - 4 and dy/du = 4 u 3

• Use the chain rule

f '(x) = (dy/du) (du/dx) = (4 u 3) (3 x 2 - 4)

• We now substitute u = x 3 - 4x + 5 above to obtain

f '(x) = 4 (x 3 - 4x + 5) 3 (3 x 2 - 4)

Example 3: f(x) = sqrt (x 2 + 2x -1)where sqrt means square root Solution to Example 3

• Let u = x 2 + 2x -1 and y = sqrt (u) , hence du/dx = 2x + 2 and dy/du = 1 / (2 sqrt(u))

• Use the chain rule f '(x) = (dy/du) (du/dx) = [ 1 / (2 sqrt(u)) ] (2x + 2)

• Substitute u = x 2 + 2x -1 above to obtain

f '(x) = (2x + 2) [ 1 / (2 sqrt(x 2 + 2x -1)) ]

• Factor 2 in numerator and denominator and simplify

f '(x) = (x + 1) / (sqrt(x

2

+ 2x -1))

Find the first derivative of f if f is given by

Example 4: f(x) = sin 2 (2x + 3)

Solution to Example 4

• Let u = sin (2x + 3) and y = u 2 , hence du/dx = 2 cos(2x + 3) and dy/du = 2 u

• Use the chain rule f '(x) = (dy/du) (du/dx) = 2 u 2 cos(2x + 3)

• Substitute u = sin (2x + 3) above to obtain f '(x) = 4 sin (2x + 3) cos (2x + 3)

• Use the trigonometric fromula sin (2x) = 2 sinx cos x to simplify f '(x) f '(x) = 2 sin (4x + 6)

Exercises: Find the first derivative to each of the functions.

1 - f(x) = cos (3x -3) (ans. f '(x) = -3 sin (3x -3) )



2 - l(x) = (3x 2 -3x + 8) 4 (ans. l(x) = 12 (2x - 1) (3x 2 -3x + 8) 3 )

3 - m(x) = sin [ 1 / (x - 2)] (ans. m(x) = -1 / (x - 2) 2 cos [ 1 / (x - 2)])

4 - t(x) = sqrt (3x 2 - 3x + 6) (ans. t(x) = 3(2x - 1) / sqrt (3x 2 - 3x + 6))

5 - r(x) = sin 2 (4x + 20) (ans. r(x) = 4 sin (8x + 4))

Solve Rate of Change Problems in Calculus

Problem 1: A rectangular water tank (see figure below) is being filled at the constant rate of 20 liters /second. The base of the tank has dimensions w = 1 meter and L = 2 meters. What is the rate of change of

the height of water in the tank?(express the answer in cm / sec).

Solution to Problem 1:

• The volume V of water in the tank is given by.

V = w*L*H

• We know the rate of change of the volume dV/dt = 20

liter /sec. We need to find the rate of change of the

height H of water dH/dt. V and H are functions of time.

We can differentiate both side of the above formula to obtain



dV/dt = W*L*dH/dt

• note W and L do not change with time and are therefore considered as constants in the above

operation of differentiation.

• We now find a formula for dH/dt as follows.

dH/dt = dV/dt / W*L

• We need to convert liters into cubic cm and meters into cm as follows

1 liter = 1 cubic decimeter

= 1000 cubic centimeters

= 1000 cm 3

and 1 meter = 100 centimeter.

• We now evaluate the rate of change of the height H of water.

dH/dt = dV/dt / W*L

= ( 20*1000 cm 3 / sec ) / (100 cm * 200 cm)

= 1 cm / sec.

Problem 2:An airplane is flying in a straight direction and at a constant height of 5000 meters (seefigure below). The angle of elevation of the airplane from a fixed point of observation is a. The speed of

the airplane is 500 km / hr. What is the rate of change of angle a when it is 25 degrees? (Express the

answer in degrees / second and round to one decimal place).


• The airplane is flying horizontally at the rate of

dx/dt = 500 km/hr. We need a relationship between

angle a and distance x. From trigonometry, we can

write

tan a = h/x

• angle a and distance x are both functions of time t.

Differentiate both sides of the above formula with

respect to t.

d(tan a)/dt = d(h/x)/dt

• We now use the chain rule to further expand the terms in the above formula

d(tan a)/dt = (sec 2 a) da/dt



d(h/x)/dt = h*(-1 / x 2) dx/dt.

(note: height h is constant)

• Substitute the above into the original formula to obtain

(sec 2 a) da/dt = h*(-1 / x 2) dx/dt

• The above can be written as

da/dt = [ h*(-1 / x 2) dx/dt ] / (sec 2 a)

• We now use the first formula to find x in terms of a and h follows

x = h / tan a

• Substitute the above into the formula for da/dt and simplify

da/dt = [ h*(- tan 2a / h 2) dx/dt ] / (sec 2 a)

= [ (- tan 2a / h) dx/dt ] / (sec 2 a)

= (- sin 2a / h) dx/dt

• Use the values for a, h and dx/dt to approximate da/dt with the right conversion of units: 1km = 1000 m and 1

hr = 3600 sec.

da/dt = [- sin 2(25 deg)/5000 m]*[500 000 m/3600 sec]

= -0.005 radians/sec

= -0.005 * [ 180 degrees / Pi radians] /sec

= -0.3 degrees/sec

Problem 3: If two resistors with resistances R1 and R2 are connected in parallel as shown in the figure

below, their electrical behavior is equivalent to a resistor of resistance R such that

1 / R = 1 / R1 + 1 / R2

If R1 changes with time at a rate r = dR1/dt and R2 is constant, express the rate of change dR / dt of the

resistance of R in terms of dR1/dt, R1 and R2.


• We start by differentiating, with respect to time, both

sides of the given formula for resistance R, noting that

R2 is constant and d(1/R2)/dt = 0

(-1/R 2)dR/dt = (-1/R1 2)dR1/dt



• Arrange the above to obtain

dR/dt = (R/R1) 2dR1/dt

• From the formula 1 / R = 1 / R1 + 1 / R2, we can write

R = R1*R2 / (R1 + R2)

• Substitute R in the formula for dR/dt and simplify

dR/dt = (R1*R2 / R1*(R1 + R2)) 2dR1/dt

= (R2 / (R1 + R2)) 2dR1/dt

Exercises

1 - Find a formula for the rate of change dV/dt of the volume of a balloon being inflated such that it radius Rincreases at a rate equal to dR/dt.

2 - Find a formula for the rate of change dA/dt of the area A of a square whose side x centimeters changes at a rate

equal to 2 cm/sec.

3 - Two cars start moving from the same point in two directions that makes 90 degrees at the constant speeds of s1

and s2. Find a formula for the rate of change of the distance D between the two cars.

solutions to the above exercises

1 - dV/dt = 4*Pi*R 2dR/dt

2 - dA/dt = 4x cm 2/sec

3 - dD/dt = sqrt( s1 2 + s2 2 )

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