Author Archives: Engineering Math

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 16

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PROBLEM:

Evaluate lim⁡x→0(1−cos⁡2(x)1+cos⁡(x)) \displaystyle \lim\limits_{x\to 0}\left(\frac{1-\cos^2\left(x\right)}{1+\cos\left(x\right)}\right)

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SOLUTION:

This problem can be solved using a direct substitution of x=0x=0. That is

lim⁡x→0(1−cos⁡2x1+cos⁡x)=1−cos⁡2(0)1+cos⁡(0)=1−11+1=0  (Answer)\begin{align*} \lim\limits_{x\to 0}\left(\frac{1-\cos^2 x }{1+\cos x }\right) & =\frac{1-\cos^2\left(0\right)}{1+\cos\left(0\right)} \\ \\ & =\frac{1-1}{1+1}\\ \\ & = 0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 15

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PROBLEM:

Evaluate lim⁡x→0(sin⁡3xsin⁡x−tan⁡x) \displaystyle \lim_{x\to 0}\left(\frac{\sin^3x}{\sin x-\tan x}\right).

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SOLUTION:

A straight substitution of x=π4\displaystyle x=\frac{\pi }{4} leads to the indeterminate form 00\displaystyle \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→0(sin⁡3xsin⁡x−tan⁡x)=lim⁡x→0(sin⁡3xsin⁡x−sin⁡xcos⁡x)=lim⁡x→0(sin⁡3xsin⁡xcos⁡x−sin⁡xcos⁡x)=lim⁡x→0(sin⁡3xcos⁡xsin⁡xcos⁡x−sin⁡x)=lim⁡x→ 0(sin⁡3xcos⁡x(sin⁡x)(cos⁡x−1))=lim⁡x→0(sin⁡2xcos⁡x(cos⁡x−1))=lim⁡x→0((1−cos⁡2x)⋅cos⁡x−(1−cos⁡x))=lim⁡x→0((1+cos⁡x)(1−cos⁡x)⋅cos⁡x−(1−cos⁡x))=lim⁡x→0((1+cos⁡x)⋅cos⁡(x)−1)=−1⋅lim⁡x→0((1+cos⁡x)⋅cos⁡x)=−1⋅(1+cos⁡0)⋅cos⁡0=−1⋅(1+1)⋅1=−2  (Answer)\begin{align*} \displaystyle \lim _{x\to 0}\left(\frac{\sin^3x}{\sin x-\tan x}\right) & =\lim _{x\to 0}\left(\frac{\sin^3x}{\sin x-\frac{\sin x}{\cos x}}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\sin^3 x}{\frac{\sin x \cos x - \sin x}{\cos x}}\right) \\ \\ & = \lim _{x\to 0}\left(\frac{\sin^3 x \cos x }{\sin x \cos x - \sin x}\right) \\ \\ &=\lim _{x\to \:0}\left(\frac{\sin^3 x \cos x}{\left(\sin x \right)\left(\cos x-1\right)}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\sin^2 x \cos x }{\left(\cos x -1\right)}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\left(1-\cos^2 x \right) \cdot\cos x }{-\left(1-\cos x \right)}\right) \\ \\ & =\lim\limits_{x\to 0}\left(\frac{\left(1+\cos x \right) \left(1-\cos x \right) \cdot\cos x }{-\left(1-\cos x \right)}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\left(1+\cos x \right) \cdot \cos\left(x\right)}{-1}\right) \\ \\ & =-1\cdot \lim _{x\to 0}\left(\left(1+\cos x\right)\cdot\cos x \right) \\ \\ & =-1\cdot \left(1+\cos 0 \right)\cdot \cos 0 \\ \\ & =-1\cdot \left(1+1\right)\cdot 1 \\ \\ & =-2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 14

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PROBLEM:

Evaluate lim⁡x→π4(tan⁡ 2xsec⁡ 2x)\displaystyle \lim\limits_{x\to \frac{\pi }{4}}\left(\frac{\tan\:2x}{\sec\:2x}\right).

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SOLUTION:

A straight substitution of x=π4x=\frac{\pi }{4} leads to the indeterminate form 00\frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→ π 4(tan⁡ 2xsec⁡ 2x)=lim⁡x→π4(sin⁡ 2xcos⁡ 2x1cos⁡ 2x)=lim⁡x→ π 4(sin⁡ 2x)=sin⁡(2⋅π4)=sin⁡π2=1  (Answer)\begin{align*} \displaystyle \lim\limits_{x\to \:\frac{\pi \:}{4}}\left(\frac{\tan\:2x}{\sec\:2x}\right) & =\lim\limits_{x\to \frac{\pi }{4}}\left(\frac{\frac{\sin\:2x}{\cos\:2x}}{\frac{1}{\cos\:2x}}\right) \\ \\ & =\lim\limits_{x\to \:\frac{\pi \:}{4}}\left(\sin\:2x\right) \\ \\ & =\sin\left(2\cdot \frac{\pi }{4}\right) \\ \\ & =\sin\frac{\pi }{2} \\ \\ & =1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 13

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PROBLEM:

Evaluate lim⁡x→3(x2−9x−3)\displaystyle \lim\limits_{x\to 3}\left(\frac{\sqrt{x^2-9}}{x-3}\right).

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SOLUTION:

A straight substitution of x=3x=3 leads to the indeterminate form 00\frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→ 3(x2−9x−3)=lim⁡x→3(x2−9x−3⋅x2−9x2−9)=lim⁡x→3((x+3)(x−3)(x−3)x2−9)=lim⁡x→3(x2−9(x−3)x2−9)=lim⁡x→3(x+3x2−9)=3+332−9=60=∞  (Answer)\begin{align*} \lim\limits_{x\to \:3}\left(\frac{\sqrt{x^2-9}}{x-3}\right) & =\lim\limits_{x\to 3}\left(\frac{\sqrt{x^2-9}}{x-3}\cdot \frac{\sqrt{x^2-9}}{\sqrt{x^2-9}}\right) \\ \\ & =\lim\limits_{x\to 3}\left(\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\sqrt{x^2-9}}\right) \\ \\ & =\lim\limits_{x\to 3}\left(\frac{x^2-9}{\left(x-3\right)\sqrt{x^2-9}}\right) \\ \\ & =\lim _{x\to 3}\left(\frac{x+3}{\sqrt{x^2-9}}\right) \\ \\ & =\frac{3+3}{\sqrt{3^2-9}} \\ \\ & =\frac{6}{0} \\ \\ & =\infty \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Since the function’s limit is different from the left to its limits from the right, the limit does not exist. 

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 12

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PROBLEM:

Evaluate lim⁡x→0(1x(13−1x+9))\displaystyle \lim\limits_{x\to 0}\left(\frac{1}{x}\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)\right)

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SOLUTION: 

A straight substitution of x=0x=0 leads to the indeterminate form 0⋅00\cdot 0 which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→ 0(1x(13−1x+9))=lim⁡x→0(1⋅ (13−1x+9)x)=lim⁡x→0(x+9−33x+9x)=lim⁡x→0(x+9−33xx+9)=lim⁡x→0(x+9−33xx+9)⋅x+9+3x+9+3=lim⁡x→0(x3x(x+9+3)x+9)=lim⁡x→0(13(x+9+3)x+9)=13(0+9+3)0+9=154  (Answer)\begin{align*} \lim\limits_{x\to \:0}\left(\frac{1}{x}\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)\right) & =\lim\limits_{x\to 0}\left(\frac{1\cdot \:\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)}{x}\right) \\ \\ & =\lim\limits_{x\to 0}\left(\frac{\frac{\sqrt{x+9}-3}{3\sqrt{x+9}}}{x}\right) \\ \\ & =\lim\limits_{x\to 0}\left(\frac{\sqrt{x+9}-3}{3x\sqrt{x+9}}\right) \\ \\ & =\lim\limits_{x\to 0}\left(\frac{\sqrt{x+9}-3}{3x\sqrt{x+9}}\right)\cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3} \\ \\ & =\lim\limits_{x\to 0}\left(\frac{x}{3x\left(\sqrt{x+9}+3\right)\sqrt{x+9}}\right) \\ \\ & =\lim\limits_{x\to 0}\left(\frac{1}{3\left(\sqrt{x+9}+3\right)\sqrt{x+9}}\right) \\ \\ & =\frac{1}{3\left(\sqrt{0+9}+3\right)\sqrt{0+9}} \\ \\ & =\frac{1}{54} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 11

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Evaluate lim⁡x→3(x−3x−2−4−x) \displaystyle \lim\limits_{x\to 3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right)

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Solution:

A straight substitution of x=3x=3 leads to the indeterminate form 00\frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows.

lim⁡x→ 3(x−3x−2−4−x)=lim⁡x→  3(x−3x−2−4−x)⋅x−2+4−xx−2+4−x=lim⁡x→3[(x−3)(x−2+4−x)(x−2−4−x)(x−2+4−x)]=lim⁡x→3[(x−3)(x−2+4−x)2x−6]=lim⁡x→3[(x−3)(x−2+−x+4)2(x−3)]=lim⁡x→3[x−2+4−x2]=3−2+4−32=1  (Answer)\begin{align*} \begin{align*} \lim\limits_{x\to \:3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right) & =\lim\limits_{x\to \:\:3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right)\cdot \frac{\sqrt{x-2}+\sqrt{4-x}}{\sqrt{x-2}+\sqrt{4-x}} \\ & =\lim\limits_{x\to 3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{\left(\sqrt{x-2}-\sqrt{4-x}\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}\right] \\ & =\lim\limits_{x\to3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{2x-6}\right]\\ & =\lim\limits_{x\to3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{-x+4}\right)}{2\left(x-3\right)}\right] \\ & =\lim\limits_{x\to 3}\left[\frac{\sqrt{x-2}+\sqrt{4-x}}{2}\right]\\ & =\frac{\sqrt{3-2}+\sqrt{4-3}}{2}\\ & =1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*} \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 10

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PROBLEM:

Evaluate lim⁡x→2(x3−8x2−4 )\displaystyle \lim\limits_{x\to 2}\left(\frac{x^3-8}{x^2-4\:}\right)

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SOLUTION:

A straight substitution of  x=2x=2 leads to the indeterminate form 00 \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→2(x3−8x2−4 )=lim⁡x→2[(x−2)(x2+2x+4)(x+2)(x−2)]=lim⁡x→2[(x2+2x+4)(x+2)]=22+2⋅2+42+2=3  (Answer)\begin{align*} \lim\limits_{x\to 2}\left(\frac{x^3-8}{x^2-4\:}\right) & =\lim\limits_{x\to 2}\left[\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x-2\right)}\right] \\ &=\lim\limits_{x\to 2}\left[\frac{\left(x^2+2x+4\right)}{\left(x+2\right)}\right] \\ & =\frac{2^2+2\cdot 2+4}{2+2} \\ & =3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 9

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PROBLEM:

Evaluate lim⁡x→4(1x−14x−4 )\displaystyle \lim\limits_{x\to 4}\left(\frac{\frac{1}{x}-\frac{1}{4}}{x-4\:}\right).

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SOLUTION:

A straight substitution of x=4x=4 leads to the indeterminate form 00\frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows:

lim⁡x→4(1x−14x−4)=lim⁡x→4(4−x4xx−4)=lim⁡x→44−x4x(x−4)=lim⁡x→4(4−x−4x(4−x))=lim⁡x→4−14x=−14⋅4=−116  (Answer)\begin{align*} \\ \lim\limits_{x\to 4}\left(\frac{\frac{1}{x}-\frac{1}{4}}{x-4}\right)& =\lim\limits_{x\to 4}\left(\frac{\frac{4-x}{4x}}{x-4}\right)\\ \\ & =\lim\limits_{x\to 4}\frac{4-x}{4x\left(x-4\right)}\\ \\ &=\lim\limits_{x\to 4}\left(\frac{4-x}{-4x\left(4-x\right)}\right)\\ \\ & =\lim\limits_{x\to 4}-\frac{1}{4x}\\ \\ & =-\frac{1}{4\cdot 4}\\ \\ & =-\frac{1}{16} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists 8th Ed | Walpole| Problem 1.2

According to the journal Chemical Engineering, an important property of a fiber is its water absorbency. A random sample of 20 pieces of cotton fiber is taken and the absorbency on each piece was measured. The following are the absorbency values:

18.71 21.41 20.72 21.81 19.29 22.43 20.17
23.71 19.44 20.50 18.92 20.33 23.00 22.85
19.25 21.77 22.11 19.77 18.04 21.12  

a. Calculate the sample mean and median for the above sample values.

b. Compute the 10% trimmed mean.

c. Do a dot plot of the absorbency data. 

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Problem 1-2|Stress | Mechanics of Materials| Ninth Edition| R.C. Hibbeler|

Determine the resultant internal normal and shear force in the member at (a) section a–a and (b) section b–b, each of which passes through point A. The 500-lb load is applied along the centroidal axis of the member.

Problem 1-2

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