Tag Archives: Feliciano and Uy

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 2

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PROBLEM:

Evaluate \displaystyle \lim\limits_{x\to 3}\left(\frac{4x+2}{x+4}\right).


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SOLUTION:

\begin{align*}

\lim_{x\to 3}\left(\frac{4x+2}{x+4}\right)& =\frac{\lim\limits_{x\to 3}\left(4x+2\right)}{\lim\limits_{x\to 3}\left(x+4\right)}\\

& =\frac{\lim\limits_{x\to 3}\left(4x\right)+\lim\limits_{x\to 3}\left(2\right)}{\lim\limits_{x\to 3}\left(x\right)+\lim\limits_{x\to 3}\left(4\right)}\\

& =\frac{4\cdot \lim\limits_{x\to 3}\left(x\right)+2}{3+4}\\

& =\frac{4\cdot 3+2}{3+4}\\

& =\frac{12+2}{7}\\

& =\frac{14}{7}\\

& =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 1

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PROBLEM:

Evaluate \displaystyle \lim _{x\to 2}\left(x^2-4x+3\right).


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SOLUTION:

\begin{align*}

\lim_{x\to 2}\left(x^2-4x+3\right)& = \lim_{x\to 2}\left(x^2\right)-\lim_{x\to 2}\left(4x\right)+\lim_{x\to 2}\left(3\right)\\

& =\left[\lim_{x\to 2}\left(x\right)\right]^2-4\lim_{x\to 2}\left(x\right)+3\\

& =\left(2\right)^2-4\left(2\right)+3\\

& =-1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 10

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PROBLEM:

If  \displaystyle f\left(x\right)=\frac{4}{x+3} and \displaystyle \:g\left(x\right)=x^2-3 , find \displaystyle f\left[g\left(x\right)\right] and \displaystyle g\left[f\left(x\right)\right].


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SOLUTION:

Part A

\begin{align*}

f\left[g\left(x\right)\right] & =\frac{4}{\left(x^2-3\right)+3}\\

& =\frac{4}{x^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}

Part B

\begin{align*}

g\left[f\left(x\right)\right] & =\left(\frac{4}{x+3}\right)^2-3\\

& =\frac{16}{\left(x+3\right)^2}-3\\

& =\frac{16-3\left(x+3\right)^2}{\left(x+3\right)^2}\\

& =\frac{16-3\left(x^2+6x+9\right)}{\left(x+3\right)^2}\\

& =\frac{16-3x^2-18x-27}{\left(x+3\right)^2}\\

& =\frac{-3x^2-18x-11}{\left(x+3\right)^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 9

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PROBLEM:

If \displaystyle f\left(x\right)=3x^2-4x+1, find \displaystyle \frac{f\left(h+3\right)-f\left(3\right)}{h},\:h\ne 0.


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SOLUTION:

\begin{align*}
\frac{f\left(h+3\right)-f\left(3\right)}{h} & =\frac{\left[3\left(h+3\right)^2-4\left(h+3\right)+1\right]-\left[3\left(3\right)^2-4\left(3\right)+1\right]}{h} \\

& =\frac{3\left(h^2+6h+9\right)-4h-12+1-16}{h}\\

& =\frac{3h^2+18h+27-4h-12+1-16}{h}\\

& =\frac{3h^2+14h}{h}\\

& =\frac{h\left(3h+14\right)}{h}\\
& =3h+14 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 8

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PROBLEM:

If \displaystyle f\left(x\right)=x^2+1, find \displaystyle \frac{f\left(x+h\right)-f\left(x\right)}{h},\:h\ne 0.


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SOLUTION:

\begin{align*}
\displaystyle \frac{f\left(x+h\right)-f\left(x\right)}{h} & =\frac{\left[\left(x+h\right)^2+1\right]-\left(x^2+1\right)\:}{h}\\ \\
& =\frac{x^2+2xh+h^2+1-x^2-1}{h}\\ \\
& =\frac{2xh+h^2}{h}\\ \\
& =\frac{h\left(2x+h\right)}{h}\\ \\
& =2x+h \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 7

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PROBLEM:

A right circular cylinder, a radius of base x, height y, is inscribed in a right circular cone, radius of base r and a height h. Express y as a function of x (r and h are constants).


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SOLUTION:

Refer to the figure below for an elevation view.

Schematic Diagram of a right circular cylinder inscribed in a right circular cone.
Diagram of a right circular cylinder with a base radius of r and height y inscribed in a right circular cone with base radius r and height h.

By ratio and proportion of two similar triangles, we have

\begin{align*}
\frac{y}{r-x} & = \frac{h}{r} \\
y & =\frac{h\left(r-x\right)\:}{r} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1 Problem 6

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PROBLEM:

The stiffness of a beam of rectangular cross-section is proportional to the breadth and the cube of the depth. If the breadth is 20 cm, express the stiffness as a function of the depth.


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SOLUTION:

Let S=stiffness, b=breadth, and d=depth

\begin{align*}
S & =bd^3 \\
S & = 20 d^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1 Problem 5

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PROBLEM:

Express the area A of an equilateral triangle as a function of its side x.


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SOLUTION:

From the formula of the area of a triangle, \displaystyle A=\frac{1}{2} \text{a}\text{b} \sin\left(\theta \right). Also, we know that the interior angle of an equilateral triangle is 60 degrees, and \displaystyle \sin\:60^{\circ} =\frac{\sqrt{3}}{2}.

\begin{align*}
A & =\frac{1}{2} \text{a}\text{b} \sin\left(\theta \right) \\
A & =\frac{1}{2} \cdot x\cdot x\cdot \sin\:60^{\circ} \\
A & =\frac{1}{2}\cdot x^2\cdot \frac{\sqrt{3}}{2} \\
A & =\frac{\sqrt{3}}{4}x^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1 Problem 4

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PROBLEM:

Express the distance D traveled in t hr by a car whose speed is 60 km/hr.


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SOLUTION:

\begin{align*}
\text{Distance} & = \text{Rate} \times \text{Time} \\
D & =\left(60\:\text{km/hr} \right)\cdot t \ \text{hr} \\
D & =60t \ \text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1 Problem 3

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PROBLEM:

If \displaystyle y= \tan\left(x+\pi \right), find x as a function of y.


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SOLUTION:

\begin{align*}
y & = \tan\left(x+\pi \right) \\
x+\pi &  = \tan^{-1}y \\
x & = \tan^{-1}y-\pi \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

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