Author Archives: Engineering Math

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 8

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PROBLEM:

Evaluate lim⁡x→0(3x+2x2−2x+4)\displaystyle \lim\limits_{x\to 0}\left(\frac{3x+2}{x^2-2x+4}\right).

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SOLUTION:

Plug in the value x=0.

lim⁡x→0(3x+2x2−2x+4)=3(0)+2(0)2−2(0)+4=0+20−0+4=24=12  (Answer)\begin{align*} \lim\limits_{x\to 0}\left(\frac{3x+2}{x^2-2x+4}\right)& =\frac{3\left(0\right)+2}{\left(0\right)^2-2\left(0\right)+4}\\ \\ & =\frac{0+2}{0-0+4}\\ \\ & =\frac{2}{4}\\ \\ & =\frac{1}{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 7

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PROBLEM:

Evaluate lim⁡x→3(3xxx+1)\displaystyle \lim\limits_{x\to 3}\left(\frac{\sqrt{3x}}{x\sqrt{x+1}}\right).

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SOLUTION:

Plug the value x=3.

lim⁡x→3(3xxx+1)=3(3)33+1=934=33⋅2=36=12  (Answer)\begin{align*} \lim\limits_{x\to 3}\left(\frac{\sqrt{3x}}{x\sqrt{x+1}}\right)&=\frac{\sqrt{3\left(3\right)}}{3\sqrt{3+1}} \\ \\ &=\frac{\sqrt{9}}{3\sqrt{4}} \\ \\ & =\frac{3}{3\cdot 2}\\ \\ & =\frac{3}{6}\\ \\ &=\frac{1}{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 6

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PROBLEM:

Evaluate lim⁡x→2(4x−3)(x2+5)\displaystyle \lim_{x\to 2}\left(4x-3\right)\left(x^2+5\right).

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SOLUTION:

Plug the value x=2.

lim⁡x→2(4x−3)(x2+5)=[(4⋅2)−3][(2)2+5]=[8−3][4+5]=(5)(9)=45  (Answer)\begin{align*} \lim\limits_{x\to 2}\left(4x-3\right)\left(x^2+5\right) & =\left[\left(4\cdot 2\right)-3\right]\left[\left(2\right)^2+5\right]\\ & =\left[8-3\right]\left[4+5\right]\\ & =\left(5\right)\left(9\right)\\ & =45 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 5

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PROBLEM:

Evaluate lim⁡x→8(2x+x3−4)\displaystyle \lim\limits_{x\to 8}\left(2x+\sqrt[3]{x}-4\right).

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SOLUTION:

Plug in the value x=8.

lim⁡x→8(2x+x3−4)=[2(8)+83−4]=[16+2−4]=14  (Answer)\begin{align*} \lim\limits_{x\to 8}\left(2x+\sqrt[3]{x}-4\right) & = \left[2\left(8\right)+\sqrt[3]{8}-4\right]\\ & =\left[16+2-4\right]\\ & =14 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 4

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PROBLEM:

Evaluate lim⁡x→π3(sin⁡ 2xsin⁡ x)\displaystyle \lim\limits _{x\to \frac{\pi }{3}}\left(\frac{\sin\:2x}{\sin\:x}\right).

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SOLUTION:

Plug in the value x=π3\displaystyle x=\frac{\pi }{3}.

lim⁡x→π3(sin⁡ 2xsin⁡ x)=sin⁡(2⋅π3)sin⁡ (π3)=3232=1  (Answer)\begin{align*} \lim\limits_{x\to \frac{\pi }{3}}\left(\frac{\sin\:2x}{\sin\:x}\right) & =\frac{\sin\left(2\cdot \frac{\pi }{3}\right)}{\sin\:\left(\frac{\pi }{3}\right)} \\ & =\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}\\ & =1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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College Physics by Openstax Chapter 2 Problem 35

Consider a grey squirrel falling out of a tree to the ground.

(a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel’s velocity just before hitting the ground, assuming it fell from a height of 3.0 m.

(b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.

Solution:

Part A

We are given the following: v0=0 m/sv_0=0 \ \text{m/s}; a=−9.80 m/s2a=-9.80 \ \text{m/s}^2; and Δx=−3.0 m\Delta x=-3.0\ \text{m}.

Note that the acceleration is due to gravity and its value is constant at a=−9.80 m/s2a=-9.80 \ \text{m/s}^2. Also, the distance xx, is negative because of the direction of motion. For free-fall, downward motion is considered negative. To solve for the velocity just before it hits the ground, we will solve vfv_fin the formula

(vf)2=(v0)2+2aΔx\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Substituting the given values:

(vf)2=(v0)2+2aΔxvf=(v0)2+2aΔxvf=(0 m/s)2+2(−9.80 m/s2)(−3.0 m)vf=7.7 m/s  (Answer)\begin{align*} \left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\ v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\ v_f & = \sqrt{\left(0\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(-3.0\:\text{m}\right)} \\ v_f & = 7.7\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are given the following: vf=0 m/sv_f=0 \ \text{m/s}; v0=7.7 m/sv_0=7.7 \ \text{m/s}; and Δx=0.02 m\Delta x=0.02 \ \text{m}

To solve for the acceleration we shall use the same formula as that in Part A. Solving for the acceleration aa:

a=(vf)2−(v0)22Δxa=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values:

a=(vf)2−(v0)22Δxa=(0 m/s)2−(7.7 m/s)22(0.02 m)a=−1.5×103 m/s2  (Answer)\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0\:\text{m/s}\right)^2-\left(7.7\:\text{m/s}\right)^2}{2\left(0.02\:\text{m}\right)} \\ a & =-1.5\times 10^3\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

This is approximately 3 times the deceleration of the airman from the previous problem, who was falling from thousands of meters high.

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College Physics by Openstax Chapter 2 Problem 34

In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few life-threatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot’s speed upon impact was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m.

Solution:

We are given the following: x=3 mx=3\ \text{m}; v0=54 m/sv_0=54\ \text{m/s}; and vf=0 m/sv_f=0 \ \text{m/s}.

We are required to solve for the acceleration, and we are going to use the formula

(vf)2=(v0)2+2aΔx\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Solving for acceleration aa in terms of the other variables:

a=(vf)2−(v0)22Δxa=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values:

a=(vf)2−(v0)22Δxa=(0 m/s)2−(54 m/s)22(3.0 m)a=−486 m/s2  (Answer)\begin{align*} a & = \frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0\:\text{m/s}\right)^2-\left(54\:\text{m/s}\right)^2}{2\left(3.0\:\text{m}\right)} \\ a & =-486\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The negative acceleration means that the pilot was decelerating at a rate of 486 m/s every second.

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College Physics by Openstax Chapter 2 Problem 33

An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m.

a) What is his deceleration?

b) How long does the collision last?

Solution:

We are given the following: v0=7.50 m/sv_0=7.50\:\text{m/s}; vf=0.00 m/sv_f=0.00\:\text{m/s}; andΔx=0.350 m\Delta x=0.350\:\text{m}.

Part A

We are going to use the formula

(vf)2=(v0)2+2aΔx\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Solving for the acceleration aa in terms of the other variables:

a=(vf)2−(v0)22Δxa=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values:

a=(vf)2−(v0)22Δxa=(0 m/s)2−(7.50 m/s)22(0.350 m)a=−80.4 m/s2  (Answer)\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0\:\text{m/s}\right)^2-\left(7.50\:\text{m/s}\right)^2}{2\left(0.350\:\text{m}\right)} \\ a & =-80.4\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are going to use the formula

Δx=vavet\Delta x=v_{ave}t

Since vave=vf+v02v_{ave}=\frac{v_f+v_0}{2}, we can write the formula as

Δx=vf+v02⋅t\Delta x=\frac{v_f+v_0}{2}\cdot t

Solving for time tt in terms of the other variables:

 t=2Δxvf+v0\:t=\frac{2\Delta x}{v_f+v_0}

Substituting the given values:

t=2Δxvf+v0t=2(0.350 m)0 m/s+7.50 m/st=9.33×10−2 s  (Answer)\begin{align*} t&=\frac{2\Delta x}{v_f+v_0} \\ t&=\frac{2\left(0.350\:\text{m}\right)}{0\:\text{m/s}+7.50\:\text{m/s}} \\ t& =9.33\times 10^{-2}\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 32

A woodpecker’s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker’s head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm.

a) Find the acceleration in m/s2 and in multiples of g (g=9.80 m/s2).

b) Calculate the stopping time.

c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain’s deceleration, expressed in multiples of g?

Solution:

We are given the following values: v0=0.600 m/sv_0=0.600 \ \text{m/s}; vf=0.000 m/sv_f=0.000\:\text{m/s}; and Δx=0.002 m\Delta x=0.002\:\text{m}.

Part A

The acceleration is computed based on the formula,

(vf)2=(v0)2+2aΔx\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Solving for acceleration aa in terms of the other variables, we have

a=(vf)2−(v0)22Δxa=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substituting the given values,

a=(vf)2−(v0)22Δxa=(0.000 m/s)2−(0.600 m/s)22(0.002 m)a=−90.0 m/s2  (Answer)\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0.000\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(0.002\:\text{m}\right)} \\ a & =-90.0\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

In terms of gg taking absolute values of the acceleration , we have

a=90.0 m/s2⋅(g9.80 m/s2)a=90.09.80ga=9.18g  (Answer)\begin{align*} a & = 90.0 \ \text{m/s}^2 \cdot \left( \frac{g}{9.80 \ \text{m/s}^2} \right) \\ a & = \frac{90.0}{9.80}g \\ a & = 9.18g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We shall use the formula

Δx=vavet\Delta x=v_{ave}t

where vavev_{ave} is the average velocity computed as

vave=v0+vf2v_{ave}=\frac{v_0+v_f}{2}

Solving for time tt in terms of the other variables, we have

t=2Δxv0+vft=\frac{2\Delta x}{v_0+v_f}

Substituting the given values, we have

t=2Δxv0+vft=2(0.002 m)0.600 m/s+0.000 m/st=6.67×10−3 s  (Answer)\begin{align*} t & =\frac{2\Delta x}{v_0+v_f} \\ t & =\frac{2\left(0.002\:\text{m}\right)}{0.600\:\text{m/s}+0.000\:\text{m/s}} \\ t & =6.67\times 10^{-3}\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

Employing the same formula we used in Part A, we have

a=(vf)2−(v0)22Δxa=(0.000 m/s)2−(0.600 m/s)22(0.0045 m)a=−40.0 m/s2\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & =\frac{\left(0.000\:\text{m/s}\right)^2-\left(0.600\:\text{m/s}\right)^2}{2\left(0.0045\:\text{m}\right)} \\ a & =-40.0\:\text{m/s}^2 \end{align*}

In terms of gg, taking absolute values of the acceleration

a=40.0 m/s2⋅(g9.80 m/s2)a=40.09.80ga=4.08g  (Answer)\begin{align*} a & = 40.0 \ \text{m/s}^2 \cdot \left( \frac{g}{9.80 \ \text{m/s}^2} \right) \\ a & =\frac{40.0}{9.80}g \\ a & = 4.08g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Mechanics of Materials: An Integrated Learning Approach 3rd Edition by Timothy A. Philpot Problem P1.2

A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip load. If the axial normal stress in the member must be limited to 18 ksi, determine the wall thickness required for the tube.

Solution:

We are given the following values:

Outside Diameter,D=2.50 inAxial Load,P=27 kipsMaximum Axial Stress,σ=18 ksiInside Diameter,d=D−2t\begin{align*} \text{Outside Diameter}, D & = 2.50\ \text{in} \\ \text{Axial Load}, P & = 27\ \text{kips} \\ \text{Maximum Axial Stress}, \sigma & = 18\ \text{ksi}\\ \text{Inside Diameter}, d & = D-2t \end{align*}

We are solving for the unknown wall thickness of the tube, tt.

From the definition of normal stress, solve for the minimum area required to support a 27-kip load without exceeding a stress of 18 ksi

σ=PAAmin=PσAmin=27 kips18 ksiAmin=1.500 in2\begin{align*} \sigma & =\frac{P}{A} \\ A_{\text{min}} & =\frac{P}{\sigma } \\ A_{\text{min}} & = \frac{27\:\text{kips}}{18\:\text{ksi}} \\ A_{\text{min}} & = 1.500\:\text{in}^2 \end{align*}

The cross-sectional area of the aluminum tube is given by

A=π4(D2−d2)A=\frac{\pi }{4}\left(D^2-d^2\right)

Set this expression equal to the minimum area and solve for the maximum inside diameter, dd

A=π4(D2−d2)A=π4[(2.50 in)2−d2]1.500 in2=π4[(2.50 in)2−d2]4(1.500 in2)=π[(2.50 in)2−d2]4(1.500 in2)π=(2.50 in)2−d2d2=(2.50 in2)−4(1.500 in2)πd=(2.50 in2)−4(1.500 in2)πd=2.08330 in\begin{align*} A & =\frac{\pi }{4}\left(D^2-d^2\right) \\ A & = \frac{\pi }{4}\left[\left(2.50\ \text{in}\right)^2-d^2\right] \\ 1.500\ \text{in}^2 & = \frac{\pi }{4}\left[\left(2.50\ \text{in}\right)^2-d^2\right] \\ 4 \left( 1.500\ \text{in}^2 \right) & = \pi \left[\left(2.50\ \text{in}\right)^2-d^2\right] \\ \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi} & = \left(2.50\ \text{in}\right)^2-d^2 \\ d^{2} & = \left( 2.50\ \text{in}^{2} \right) - \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi} \\ d & = \sqrt{\left( 2.50\ \text{in}^{2} \right) - \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi}} \\ d & = 2.08330\ \text{in} \end{align*}

The outside diameter DD, the inside diameter dd, and the wall thickness tt are related by

D=d+2tD = d+2t

Therefore, the minimum wall thickness required for the aluminum tube is

t=D−d2t=2.50 in−2.08330 in2t=0.208 in  (Answer)\begin{align*} t & =\frac{D-d}{2} \\ t & = \frac{2.50\:\text{in}-2.08330\:\text{in}}{2} \\ t & = 0.208\ \text{in} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}