Tutorials for the FE Examination Mathematics section

Tutorials for the FE ExaminationMathematics sectionLakshmi N. Sankarlsankar@ae.gatech.edu404-894-3014Examination covers..Solution of quadratic equationStraight linesConic sections (Parabola, Hyperbola, ellipse, circle)Spherical and ellipsoid surfaces, pyramids, polygonsLogarithmsVectorsDerivatives, integrals, ODEs and Laplace TransformsNumerical methods: root finding, derivatives, and integralsProbability and Statistics basic concepts

Some TipsBudget the time 140 questions in 4 hours, 20 of these in the math area (30 to 35 minutes).Limited time to answer 20 questionsWork backwards if that is easier, instead of forward.Plug in the supplied info into the multiple choices to see which one makes sense.Example 1An equation of a straight line through the point (6,2) with a slope of 3 isy=3x+16y=3x+20y=3x-16x=3y-16x=y/3+16

Example 2Ellipses, parabolas and hyperbolas are all examples of Close curvesCircular segmentsParabolic sectionsElliptical sectionsConical sectionsParabola is not closedThere is no such thing called parabolicOr elliptical section.

Ans: (e) Conical sectionsExample 3The partial derivative y/ x of y=x2z+3z2x+6(x+z) is2xz+3z2+6X2z+6zx+6z2x+92x+6zx+6(x+1)2x+6z+6Working forward is easiestin this case.

Ans: (a)Example 4If bacterial cells divide themselves into two every 40 minutes, the total number of cells in 240 minutes is:612243264Ans: in 40 minutes, 2 times 2= 4 In 80 minutes, 4 x 2= 8Ans: (e) 64Example 5A bag contains 100 balls numbered 1 to 100. One ball is removed. What is the probability that the number on this ball is odd or greater than 80?0.10.50.60.70.8We have 20 balls with number greater than 80.Among remaining balls numbered 1 to 80, forty are odd.

60 balls out of 100 meet this criterion.

Probability = 60/100=0.6

Ans: (c) Example 6Consider a f(x) equal to the determinant below.

The first derivative w.r.t. x is3x2-8x43x2+7x64x3-6x5

It is easiest to work forward.

Determinant is x4-x6

Take derivative

Answer: (c)Example 7If a function y=6x2 is integrated between x=-2 and x=+3, the answer is most nearly3038547078Integrate y= 6x2 first

Answer is 2x3

Apply limits.

2 (33) 2 (-2)3 = 54+16 = 70

Ans: (d)Example 8Which of the following trigonometric identities is not valid?

If you remember the relationshipsyou may immediately select (d)

Otherwise, try plugging in numerical values

For example if A=45 degrees, B is 45 degrees

We expect cos(A+B) to be zero. We do notget this from equation (d)

Thus equation (d) is not valid.Example 9The Laplace transform of a step function of magnitude a is1/(s+a)a/sa/(s+a)a/s2s+aThe Laplace transform of a unitstep function is 1/s

Multiply by height a to get a/s

Solution: (b)

Example 10One wishes to estimate the mean M of a population from a sample n drawn from this population. For the sample, the mean is x, and the standard deviation is s. The probable accuracy of the estimate increase with increase inMnXSM-sWe can eliminate X and ssince these will likely have units.

M-s will also not make sense Since M is an integer, and smay have units.

We can not increase M withoutChanging the mean value

Thus, increasing the sample sizeWill give us the best estimates.

Ans (b)Example 11If a functional form of a curve is known, differentiation may be used to determine all of the following except theSlope of the curveConcavity of the curveLocation of the inflection pointsNumber of inflection pointsArea under the curve within bounds

To find area under a curve,we integrate , not differentiate.

Thus, differentiation will not givearea.

Ans: (e)Example 12Given: dy/dt+5y=0 and y(0)=1which of the followingsolution satisfies the ODE and boundary condition?

We may use direct approachdy/y=-5dtIntegrate: logy=-5t+CApply boundary condition y(0)=1Thus C=0log y=-5ty=exp(-5t)

Ans: (b)

You may also use trial and errorFirst plug t=0 and see if y=1 toEliminate choices (d) and (e)..Example 13The general equation of a second degree polynomial is Ax2+Bxy+Cy2+Dx+Ey+F=0.

The values of the coefficients above for the parabola on the right isB=C=D=F=0, A >0, E>0B=C=D=F=0, A 0A=B=E=F=0, C >0, D>0A=B=E=F=0, D 0A=D=E=F=0, B 0xyFunction has two values of y for a given x

We need y2 term to support this.

This eliminates (a), (b)

(e) Gives a linear curve, not suitable

Try the remaining functions ( c) and (d) by setting A, B, E, F to 0

Cy2+DX=0 : Since x >0, D/C must be ve

Only possible solution is (d) Example 14If D is a differential operator, then the general solution (D-2)3y=0 isCebx (A+B+C)dy/dx(Asin2x+Bsin2x)3(A+B+C) ebx (A+Bx+Cx2) e2x The easiest way may be to reason as follows.

If we plug in a solution y= exp(bx) into (D-2)3y=0, we will get three repeating roots, with b equal to 2.

When we have three repeating roots, the solutionis of form given by (e).Example 15A particle travels on a straight line. The distance traveled s=20t3-t4. The rate of change of acceleration a time t=2 is-1402072144First derivative ds/dt gives velocity

ds/dt=60t2-4t3

Second derivative is acceleration

d2s/dt2=120t-12t2

Rate of change of acceleration is

d3s/dt3=120-24t

Plug in t=2 to get 72.

Answer is (d)

Example 16The slope of a line that is perpendicular to the line through (2,0) and (4,2) is421-1-2Find slope of given line first.

The end values of y- are 2 and 0

The end values of x are 4 and 2

Slope = change in y/change in x

Slope = (2-0)/(4-2)=1

Product of slopes of two perpendicular Lines is -1.

Answer: (d)Example 17Which of the following is a unit vector perpendicular to the plane determined by A=2i+4j and B=i+j-k?-2i+j+k1/31/2 (i+j+k)1/51/2 (i-2j)1/61/2 (-2i+j-k)1/61/2 (-2i-j-k)

The easiest approach may be tofirst take the cross product of A and B which lie on the plane.

The resulting vector is normal to A and B

Take unit vector of the result, by dividingThe vector by its magnitude.

Cross product of A and B:

-4 i +2 j -2 k

Magnitude is square root of 24, equals 2 times square root of 6

Answer (d)Example 18Consider two vectors A=2i+3j-2k and B=5i-2j+k. Find the cross product A x B.-2i+18j-6k-i-12j-19k7i+j-k10-6j-2k4

The answer is (b)Example 19If f denotes the derivative of a function f, then f(x) is defined as

From basic calculus, solution is (d)Example 20What is the area of the region in the first quadrant bounded by line y=1, x=y3/2 and the y- axis?2/53/52/31 Plot the curve with y- as the horizontal axisAnd x- as the vertical axis.yxY=1Integrate the curve from y=0 to y=1

Example 21

Assume a trial solution x= C elt

Plug into ODE. We get l= + 4i and -4i

exp(4it) and exp(-4it) both may beExpressed in terms of cos(4t) and sin(4t) terms.

Thus (e) is the general solution, beforeWe apply the initial conditions.Example 22In example 21, the solution that fits the initial condition isX=1/4e-ltX=1/3 sin(4t)X= sin(4t)+ cos(4t)X=4cos(4t)X= cos(4t)Does not match example 21 and (d) do not satisfy the initial condition x=1/4(c)does not satisfy dx/dt=0 at t=0

Only (e) satisfies ODE and both initialconditions.

Ans: (e)Example 23In example 21, what is the amplitude of motion? feet1/3 feet1 ft2 ft4 ftFrom previous slide,X=1/4cos(4t)Maximum value of x is .

Ans: (a) Example 24The period of motion for previous example isp/3 sec p/2 sec p sec2p sec3p secSin(4t) goes from 0 to 2p over aPeriod T.

4T = 2p

T= p/2

Ans: (b)Example 25Consider a curve Y(x)=8-2x2. The line tangent to this curve at (1,6) intersects x-axis at -2327/35/210The tangent line has shapey=mx+c

m=line slope= same as curve slopem= (-4x) at x=1m=-4

y=-4x+c

At x=1, y=6, thus c=10

y=-4x+10

At x-axis intersection, y=0x=10/4=5/2

Ans (d) Example 26The equation of the line normal to the point (1,6) is(y-6)/(x-1)=4(y-6)/(x-1)=1/4(y-6)/(x-1)=-4(y+6)/(x+1)=-4(x-1)/(y-6)=1/4

Consider a general point (x,y) on theNormal line, which also passes through(1,6).

Slope of this line then change in y-value divided by change in x-value

Slope= (y-6)/(x-1)

This slope will be -1/(slope of tangent)

From previous slide, slope of tangent -4

Slope of normal is +1/4

Ans: (b) Example 27Consider an area bound by y=x2, y=0 and x=4.What is the area?3/64864/343/232Integrate y with respect to x

Integral x3/3

Set the limits x=0 and x=4

Integral = [43/3- 03/3]=64/3

Ans: ( c)

Example 28What is the first moment of the above area about the x-axis?424/564/3101512/5To find the moment about the x-axis,

We divide area into slivers of height dy

The length: 4-x = 1-y1/2

Area of sliver = (4-y1/2)dy

Moment arm is y

Integrate (4-y1/2)ydy from 0 to 16

Ans: (2y2 - 2/5 y5/2) at y=4

Ans: 512-2048/5=512/5

Ans: (e)Example 29What is the volume generated by revolving this area about the x- axis?512p/5512p/3200p1024p/51024p/3

To find the volume, we use anapproach similar to example 28.

Instead of the moment arm y,

We use circumference 2py

Thus, the result in example 28is multiplied by 2p.

Ans: (d)Example 30Consider the function y=2x3-3x2-36x+25. What is the slope of the line tangent to this curve at the point (1,-12)?-42-36-172836Find slope by taking derivative.dy/dx=6x2-6x-36

Use x=1

dy/dx = 6-6-36= -36 at x=1

Ans: (b)Example 31In the previous example, at how many points will the tangent to the curve y(x) be horizontal?At no pointAt exactly one pointAt exactly two pointsAt exactly three pointsAt exactly four pointsFind slope by taking derivative.dy/dx=6x2-6x-36

Set slope to zero. Common factor 6

x2-x-6=0

Quadratic, likely two roots

(x-3)(x+2)=0

Roots are 3, -2

Ans: (c)

Example 32The function assumes a relative minimum at what value of x?-10123Since slope is zero at x=3 and x=-2

(e) Is the answer.Example 33The graph of the function in the previous example has an inflexion point at x equal to-203NoneAt the inflexion point, second derivative is zero.

Set 12x-6 to zero.

Ans (c )Example 34In the previous example, the triple derivative at x= 1 is-60612undefinedThe third derivative of

y=2x3-3x2-36x+25

is 12

Ans: (d)Example 35Consider a spring-mass-damper system d2x/dt2+5 dx/dt+6x=0

x=1/2 meter at t=0, and velocity dx/dt is 0 at t=0

What is the general solution of this equation?

Assume a solution x=elt

Plug in.

A quadratic for l results.

l2+5l+6=0

Has two roots, l= -2, -3

Two solutions that may be superposed.Ans: (b)Example 36The solution x=f(t) that fits the initial condition ise-2t/2e-3t/2-e-2t+ 3/2 e-3te-2t+ 1/2 e-3t3/2 e-2t - e-3t

Start with x = Ae-2t+ B e-3t

At t=0, x=/12 A+B=1/2

At t=0, dx/dt=0 -2A-3B=0

B= -1 and A= 3/2

Ans: (e)Example 37The displacement at t=1 ise-2(1-e-1)e-2(3/2-e-1)e-2(1- 3/2 e-1)e2(1-e-1)e2(3/2-e-1)Use result from previous example.

x = 3/2e-2t - e-3t

Ans: (b)

Example 38Consider a triangle with three vertices A (0,-2), B (1,5), and C (9,1). A line BD is drawn perpendicular to the line AC.What is the length of AC?2 times square root of 33 times square root of 10101112Find square root of [(9-0)2+(1+2)2]

Ans: (b)Example 39What is the equation of line AC?y=1/3 x -2y = -1/3 x +2x+3=6x-3y+6=0x+y=6End points of A: (0,-2)End points of C (9,1)

Slope: change in y divided by change in x

Slope of AC is (1+2)/9 = 1/3

Y= 1/3 x + c

At x=0, y=2 Thus C = -2

Y = 1/3 x 2

Ans: (a)Example 40What is the slope of line BD, perpendicular to line AC?-5-3-1/31/33From previous example, the slope ofline AC is 1/3

BD is perpendicular to AC.

Product of slopes is -1

Slope of BD is -3

Ans: (b)