FE Civil Mathematics & Computational Tools practice problems

The point-to-line distance formula is $d = |Ax_0 + By_0 + C| / \sqrt{A^2 + B^2}$. Plug in: numerator $= |3(8) - 4(6) + 10| = |24 - 24 + 10| = 10$; denominator $= \sqrt{9 + 16} = 5$; so $d = 10/5 = 2$. The value 10 is the numerator alone, from forgetting to divide by the magnitude of the coefficient vector. The value 1.4 comes from dividing by $\sqrt{2}$ instead of $\sqrt{A^2+B^2}$. The value 50 forgets the square root in the denominator.

1. Identify $A = 3$, $B = -4$, $C = 10$ and the point $(x_0, y_0) = (8, 6)$.
2. Apply the distance formula:
   $d = \frac{|3(8) - 4(6) + 10|}{\sqrt{3^2 + (-4)^2}}$
3. Compute:
   $d = \frac{|24 - 24 + 10|}{\sqrt{25}} = \frac{10}{5} = 2$

FE Reference Handbook 10.6, p. 37

The vertex of a parabola occurs at $x = -b/(2a) = 12/4 = 3$. The minimum elevation is the $y$-value there: $y = 2(3)^2 - 12(3) + 5 = 18 - 36 + 5 = -13$. The value 3 is the vertex $x$-coordinate, not the elevation, a classic mix-up. The value 5 is the $y$-intercept (the value at $x = 0$), not the minimum. The value -7 is just a distractor; computing $y$ correctly at $x=3$ gives $18-36+5=-13$ (dropping the constant term would instead give -18).

1. Locate the vertex:
   $x = -\frac{b}{2a} = -\frac{-12}{2(2)} = 3$
2. Substitute back to find the minimum elevation:
   $y = 2(3)^2 - 12(3) + 5 = 18 - 36 + 5 = -13$

FE Reference Handbook 10.6, p. 37

The quotient rule turns a difference of logs into the log of a quotient: $\log_2(48) - \log_2(3) = \log_2(48/3) = \log_2(16)$. Since $2^4 = 16$, the answer is 4. The choice $\log_2(45)$ is the result of wrongly subtracting the arguments ($48 - 3 = 45$) inside the log, but there is no such identity. The value 45 is just that wrong subtraction. The value 16 is the argument itself, forgetting to take the log.

1. Apply the quotient rule:
   $\log_2(48) - \log_2(3) = \log_2\!\left(\frac{48}{3}\right) = \log_2(16)$
2. Evaluate the logarithm:
   $\log_2(16) = 4 \quad (\text{since } 2^4 = 16)$

FE Reference Handbook 10.6, p. 38

Set $N/N_0 = 0.25 = (1/2)^{t/8}$. Since $0.25 = (1/2)^2$, the exponents must match: $t/8 = 2$, so $t = 16$ days. You can also reason it out: 25 percent is two half-lives, and each half-life is 8 days, so $2 \times 8 = 16$. The value 12 averages 8 and 16 incorrectly. The value 32 is four half-lives (which gives 6.25 percent). The value 4 is only half of one half-life.

1. Set up the decay equation:
   $0.25 = \left(\frac{1}{2}\right)^{t/8}$
2. Express 0.25 as a power of one-half:
   $\left(\frac{1}{2}\right)^{2} = \left(\frac{1}{2}\right)^{t/8}$
3. Match exponents and solve:
   $\frac{t}{8} = 2 \implies t = 16\,\text{days}$

FE Reference Handbook 10.6, p. 335

The horizontal distance (60 m) is adjacent to the angle, and the tower height is opposite. Opposite over adjacent is tangent: $\tan 32° = h/60$, so $h = 60\tan 32° = 60(0.6249) = 37.5$ m. The value 31.8 uses sine of the distance ($60\sin 32°$), 50.9 uses cosine, and 96.0 divides instead of multiplying ($60/\tan 32°$).

1. The 60 m distance is adjacent to the angle; the height is opposite. Use tangent:
   $\tan 32° = \frac{h}{60}$
2. Solve for h:
   $h = 60\tan 32° = 60(0.6249) = 37.5\,\text{m}$

FE Reference Handbook 10.6, p. 39

The angle at the horizontal leg has the 9 m vertical leg opposite it and the 12 m horizontal leg adjacent. So $\tan\theta = 9/12 = 0.75$ and $\theta = \tan^{-1}(0.75) = 36.9°$. The value 53.1 is the complementary angle (what the hypotenuse makes with the vertical leg), computed by swapping opposite and adjacent. The value 48.6 uses $\sin^{-1}(9/12)$ by mistake, and 41.4 uses $\cos^{-1}(9/12)$.

1. At the horizontal leg, the 9 m leg is opposite and the 12 m leg is adjacent:
   $\tan\theta = \frac{9}{12} = 0.75$
2. Take the inverse tangent:
   $\theta = \tan^{-1}(0.75) = 36.9°$

FE Reference Handbook 10.6, p. 39

You have a complete side-angle pair ($a$, $A$) and another side ($b$), so Law of Sines solves for angle $B$: $\sin B = b\sin A / a = 14\sin 40° / 18 = 14(0.6428)/18 = 0.4999$, giving $B = \sin^{-1}(0.5) = 30.0°$. The smaller side ($14 < 18$) must face the smaller angle ($30° < 40°$), which is a good sanity check. The value 59.4 flips the ratio ($a\sin A / b$). The value 52.7 drops the sine on angle $A$.

1. Set up the Law of Sines:
   $\frac{a}{\sin A} = \frac{b}{\sin B}$
2. Solve for $\sin B$:
   $\sin B = \frac{b\sin A}{a} = \frac{14\sin 40°}{18} = \frac{14(0.6428)}{18} = 0.500$
3. Take the inverse sine:
   $B = \sin^{-1}(0.500) = 30.0°$

FE Reference Handbook 10.6, p. 39

The largest angle is always opposite the longest side (20 m), so call it $C$. Rearrange the Law of Cosines: $\cos C = (a^2 + b^2 - c^2)/(2ab) = (121 + 169 - 400)/(2 \cdot 11 \cdot 13) = -110/286 = -0.3846$. A negative cosine means an obtuse angle: $C = \cos^{-1}(-0.3846) = 112.6°$. The value 67.4 is the supplement (forgetting the cosine is negative). The value 90.0 assumes a right angle. The value 98.2 drops the negative sign in the numerator partway through.

1. The largest angle $C$ is opposite the longest side $c = 20$. Use $a = 11$, $b = 13$:
   $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
2. Substitute:
   $\cos C = \frac{121 + 169 - 400}{2(11)(13)} = \frac{-110}{286} = -0.3846$
3. Take the inverse cosine (negative cosine gives an obtuse angle):
   $C = \cos^{-1}(-0.3846) = 112.6°$

FE Reference Handbook 10.6, p. 39

The angle 225 degrees lands in Quadrant III, with reference angle $225° - 180° = 45°$. The reference value is $\tan 45° = 1$. In Q3 both sine and cosine are negative, and tangent (sine over cosine) is positive, so $\tan 225° = +1$. The value -1 wrongly makes it negative. The value $\sqrt{3}$ is $\tan 60°$. The value $\sqrt{2}/2$ is $\sin 45°$, not its tangent.

1. Reference angle:
   $225° - 180° = 45°$
2. In Q3, both sine and cosine are negative, so tangent is positive:
   $\tan 225° = +\tan 45° = 1$

FE Reference Handbook 10.6, p. 39

Use the Pythagorean identity to get cosine: $\cos^2\theta = 1 - (8/17)^2 = 1 - 64/289 = 225/289$, so $\cos\theta = \pm 15/17$. In Quadrant II cosine is negative, so $\cos\theta = -15/17$. Then $\tan\theta = \sin\theta/\cos\theta = (8/17)/(-15/17) = -8/15$. The value +8/15 ignores the quadrant and keeps it positive. The value -15/8 inverts the ratio (cos over sin). The value 8/17 just repeats the sine.

1. Find cosine using the identity:
   $\cos^2\theta = 1 - \left(\tfrac{8}{17}\right)^2 = 1 - \tfrac{64}{289} = \tfrac{225}{289}$
2. In Q2, cosine is negative:
   $\cos\theta = -\frac{15}{17}$
3. Compute the tangent:
   $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{8/17}{-15/17} = -\frac{8}{15}$

FE Reference Handbook 10.6, p. 40

A vertical parabola in the form $(x - h)^2 = 4p(y - k)$ has vertex $(h, k)$ and focus $(h, k + p)$. Here $4p = 8$, so $p = 2$, and the vertex is (3, 2). The focus sits $p = 2$ units above the vertex: $(3, 2 + 2) = (3, 4)$. The point (3, 2) is the vertex itself. The point (3, 10) uses $p = 8$ (forgetting to divide by 4). The point (3, 6) uses $p = 4$ (dividing by 2 instead of 4).

1. Compare to standard form $(x-h)^2 = 4p(y-k)$: vertex is (3, 2) and $4p = 8$.
2. Solve for the focal distance:
   $4p = 8 \implies p = 2$
3. The parabola opens upward, so the focus is $p$ units above the vertex:
   $(3,\, 2 + 2) = (3, 4)$

FE Reference Handbook 10.6, p. 45

For an ellipse, the focal distance satisfies $c^2 = a^2 - b^2$, where $a^2$ is the larger denominator. Here $a^2 = 25$ and $b^2 = 9$, so $c^2 = 25 - 9 = 16$ and $c = 4$. The value $\sqrt{34}$ adds the denominators instead of subtracting ($\sqrt{25 + 9}$), which is the hyperbola relationship. The value 8 gives 2c (the distance between the two foci). The value 16 reports $c^2$ without taking the square root.

1. Identify the larger denominator: $a^2 = 25$, $b^2 = 9$.
2. Apply the ellipse focal relationship:
   $c^2 = a^2 - b^2 = 25 - 9 = 16$
3. Take the square root:
   $c = \sqrt{16} = 4$

FE Reference Handbook 10.6, p. 45

The power rule combined with the chain rule: bring the 4 down, reduce the exponent to 3, then multiply by the derivative of the inside (6x). That gives $4(3x^2 + 2)^3 \cdot 6x = 24x(3x^2 + 2)^3$. The form $4(3x^2+2)^3$ forgets the chain rule (the inner derivative). The form $12x(3x^2+2)^3$ uses an inner derivative of 3x instead of 6x. The form $4(6x)^3$ wrongly raises the inner derivative to the third power.

1. Apply the chain rule with outer power 4 and inner $u = 3x^2 + 2$:
   $f'(x) = 4(3x^2 + 2)^3 \cdot \frac{d}{dx}(3x^2 + 2)$
2. The inner derivative is 6x:
   $f'(x) = 4(3x^2 + 2)^3 \cdot 6x = 24x(3x^2 + 2)^3$

FE Reference Handbook 10.6, p. 50

Write the root as a power: $f(x) = (4x+1)^{1/2}$. The chain rule gives $f'(x) = \tfrac{1}{2}(4x+1)^{-1/2} \cdot 4 = 2/\sqrt{4x+1}$. At $x = 2$: $\sqrt{4(2)+1} = \sqrt{9} = 3$, so $f'(2) = 2/3$. The value 1/6 forgets the chain-rule factor of 4 (giving $1/(2\sqrt{9})$). The value 1/3 drops the factor of 4 differently. The value 3 is $\sqrt{4x+1}$ itself at $x = 2$, not its derivative.

1. Rewrite and differentiate with the chain rule:
   $f'(x) = \frac{1}{2}(4x+1)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4x+1}}$
2. Evaluate at $x = 2$:
   $f'(2) = \frac{2}{\sqrt{4(2)+1}} = \frac{2}{\sqrt{9}} = \frac{2}{3}$

FE Reference Handbook 10.6, p. 50

Acceleration is the second derivative of position. First derivative (velocity): $v(t) = 3t^2 - 12t$. Second derivative (acceleration): $a(t) = 6t - 12$. At $t = 3$: $a(3) = 18 - 12 = 6$ m/s squared. The value -9 is the velocity at $t = 3$ ($v(3) = 27 - 36 = -9$), the classic mistake of stopping at the first derivative. The value 0 is the velocity at the turning point, and 18 is 6t without the constant.

1. Velocity is the first derivative:
   $v(t) = 3t^2 - 12t$
2. Acceleration is the second derivative:
   $a(t) = 6t - 12$
3. Evaluate at $t = 3$:
   $a(3) = 6(3) - 12 = 6\,\text{m/s}^2$

FE Reference Handbook 10.6, p. 102

Let $x$ be the length of the two sides perpendicular to the channel and $y$ the side parallel to it. The fence covers three sides: $2x + y = 240$, so $y = 240 - 2x$. Area $A = xy = x(240 - 2x) = 240x - 2x^2$. Maximize: $A' = 240 - 4x = 0$ gives $x = 60$, then $y = 240 - 120 = 120$, and $A = 60 \times 120 = 7{,}200$ m squared. The value 3,600 uses a square assumption. The value 14,400 ignores that one side is open (treats it as $120 \times 120$). The value 4,800 miscomputes the dimensions.

1. Constraint with one side open:
   $2x + y = 240 \implies y = 240 - 2x$
2. Area as a function of x:
   $A(x) = x(240 - 2x) = 240x - 2x^2$
3. Maximize by setting the derivative to zero:
   $A'(x) = 240 - 4x = 0 \implies x = 60$
4. Find the dimensions and area:
   $y = 240 - 120 = 120, \quad A = 60 \times 120 = 7{,}200\,\text{m}^2$

FE Reference Handbook 10.6, p. 47

The antiderivative of $\sin x$ is $-\cos x$. Evaluate from 0 to $\pi/2$: $[-\cos x]_0^{\pi/2} = -\cos(\pi/2) - (-\cos 0) = -0 + 1 = 1$. The value -1 drops the negative sign on the antiderivative (using $+\cos x$). The value 0 comes from evaluating only at one limit. The value $\pi/2$ confuses the integral of sine with the width of the interval.

1. Find the antiderivative:
   $\int \sin x\,dx = -\cos x$
2. Evaluate at the limits:
   $\left[-\cos x\right]_0^{\pi/2} = -\cos\tfrac{\pi}{2} - (-\cos 0) = 0 + 1 = 1$

FE Reference Handbook 10.6, p. 51

First find where the curves cross: $2x = x^2$ gives $x = 0$ and $x = 2$. Between them the line 2x is on top, so integrate (top minus bottom): $\int_0^2 (2x - x^2)\,dx = [x^2 - x^3/3]_0^2 = 4 - 8/3 = 4/3$. The value 8/3 integrates $x^2$ alone (the area under the parabola, $\int_0^2 x^2\,dx = 8/3$). The value 4 is the area under the line only without subtracting ($\int_0^2 2x\,dx = 4$). The value 2/3 is a miscalculation distractor.

1. Find the intersection points:
   $2x = x^2 \implies x = 0,\ x = 2$
2. Integrate the top curve minus the bottom curve:
   $\int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2$
3. Evaluate:
   $\left(4 - \frac{8}{3}\right) - 0 = \frac{4}{3}$

FE Reference Handbook 10.6, p. 51

Direct substitution gives $(1 - 1)/0 = 0/0$, an indeterminate form, so apply L'Hopital's. Differentiate top and bottom: the top becomes $2e^{2x}$ and the bottom becomes $3\cos 3x$. Now substitute $x = 0$: $2e^0 / (3\cos 0) = 2(1)/(3(1)) = 2/3$. The value 3/2 inverts the ratio. The value 1 forgets the chain-rule factors of 2 and 3. The value 0 wrongly evaluates before differentiating.

1. Direct substitution gives 0/0. Apply L'Hopital's:
   $\lim_{x \to 0}\frac{2e^{2x}}{3\cos 3x}$
2. Evaluate at $x = 0$:
   $\frac{2e^0}{3\cos 0} = \frac{2(1)}{3(1)} = \frac{2}{3}$

FE Reference Handbook 10.6, p. 49

As $x \to \infty$ both top and bottom go to infinity, so it is $\infty/\infty$ and L'Hopital's applies. Derivative of $\ln x$ is 1/x; derivative of $\sqrt{x} = x^{1/2}$ is $\tfrac{1}{2}x^{-1/2}$. The ratio is $(1/x)/(\tfrac{1}{2}x^{-1/2}) = \tfrac{2}{\sqrt{x}}$, which goes to 0 as $x \to \infty$. Any positive power of $x$ eventually beats $\ln x$. The value $\infty$ assumes the log wins (it does not). The value 1/2 is the leftover coefficient without the limit.

1. Form is $\infty/\infty$. Apply L'Hopital's:
   $\lim_{x \to \infty}\frac{1/x}{\tfrac{1}{2}x^{-1/2}}$
2. Simplify the complex fraction:
   $= \lim_{x \to \infty}\frac{2}{\sqrt{x}}$
3. Evaluate the limit:
   $\frac{2}{\sqrt{\infty}} = 0$

FE Reference Handbook 10.6, p. 49

A direction cosine is a component divided by the vector magnitude: $\cos\beta = A_y / |\vec{A}|$. The magnitude is $\sqrt{4^2 + (-4)^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6$. So $\cos\beta = -4/6 = -2/3$. The negative sign carries through because $A_y$ is negative. The value +2/3 drops that sign. The value -4/9 divides by the squared magnitude (36) by mistake. The value 1/3 uses the $z$-component instead of the $y$-component.

1. Compute the magnitude:
   $|\vec{A}| = \sqrt{4^2 + (-4)^2 + 2^2} = \sqrt{36} = 6$
2. Direction cosine with the y-axis uses the y-component:
   $\cos\beta = \frac{A_y}{|\vec{A}|} = \frac{-4}{6} = -\frac{2}{3}$

FE Reference Handbook 10.6, p. 60

Add the vectors component by component first: $\vec{A} + \vec{B} = (3+1)\hat{i} + (-2+4)\hat{j} + (6-2)\hat{k} = 4\hat{i} + 2\hat{j} + 4\hat{k}$. Then take the magnitude: $\sqrt{4^2 + 2^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$ N. The value 10 adds the individual magnitudes ($|\vec{A}| = 7$, $|\vec{B}| \approx 4.6$), but magnitudes do not add directly. The value 36 forgets the square root. The value 8.5 adds magnitudes another way.

1. Add the vectors component by component:
   $\vec{A} + \vec{B} = 4\hat{i} + 2\hat{j} + 4\hat{k}$
2. Take the magnitude of the resultant:
   $|\vec{A} + \vec{B}| = \sqrt{4^2 + 2^2 + 4^2} = \sqrt{36} = 6\,\text{N}$

FE Reference Handbook 10.6, p. 60

Work is the dot product of force and displacement: $W = \vec{F} \cdot \vec{d} = (40)(2) + (-20)(3) + (10)(-1) = 80 - 60 - 10 = 10$ J. The negative products from the opposing components reduce the total. The value 130 mishandles the signs. The value 150 keeps everything positive (80 + 60 + 10). The value 80 uses only the first term.

1. Apply the dot product:
   $W = \vec{F} \cdot \vec{d} = (40)(2) + (-20)(3) + (10)(-1)$
2. Compute each term and sum:
   $W = 80 - 60 - 10 = 10\,\text{J}$

FE Reference Handbook 10.6, p. 61

The vector projection is $\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\vec{b}$. The dot product is $(3)(2) + (3)(-1) + (3)(2) = 6 - 3 + 6 = 9$. The squared magnitude is $|\vec{b}|^2 = 2^2 + (-1)^2 + 2^2 = 9$. So the scalar coefficient is $9/9 = 1$, giving $1 \cdot \vec{b} = 2\hat{i} - 1\hat{j} + 2\hat{k}$. The choice $6\hat{i}-3\hat{j}+6\hat{k}$ forgets to divide and just scales $\vec{b}$ by 3. The choice $3\hat{i}+3\hat{j}+3\hat{k}$ is just $\vec{a}$. The last choice divides by $3^3 = 27$ instead of $|\vec{b}|^2 = 9$.

1. Compute the dot product:
   $\vec{a}\cdot\vec{b} = (3)(2) + (3)(-1) + (3)(2) = 9$
2. Compute the squared magnitude of b:
   $|\vec{b}|^2 = 2^2 + (-1)^2 + 2^2 = 9$
3. Form the projection:
   $\text{proj}_{\vec{b}}\vec{a} = \frac{9}{9}\,\vec{b} = 2\hat{i} - 1\hat{j} + 2\hat{k}$

FE Reference Handbook 10.6, p. 61

Set up the determinant with $\hat{i}, \hat{j}, \hat{k}$ on top. The $\hat{i}$ term: $(2)(6) - (3)(5) = 12 - 15 = -3$. The $\hat{j}$ term: $-[(1)(6) - (3)(4)] = -[6 - 12] = +6$. The $\hat{k}$ term: $(1)(5) - (2)(4) = 5 - 8 = -3$. Result: $-3\hat{i} + 6\hat{j} - 3\hat{k}$. The reversed-sign choice is $\vec{B} \times \vec{A}$. The choice $-3\hat{i}-6\hat{j}-3\hat{k}$ forgets the negative on the $\hat{j}$ cofactor. The choice $4\hat{i}+10\hat{j}+18\hat{k}$ multiplies components directly (not a cross product).

1. Set up the determinant:
   $\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix}$
2. Expand each cofactor:
   $\hat{i}[(2)(6)-(3)(5)] - \hat{j}[(1)(6)-(3)(4)] + \hat{k}[(1)(5)-(2)(4)]$
3. Compute:
   $= -3\hat{i} + 6\hat{j} - 3\hat{k}$