Tag Archives: Differential and Integral Calculus

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 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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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 2

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PROBLEM:

If \displaystyle y=\frac{x^2+3}{x}, find x as a function of y.

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SOLUTION:

\begin{align*}
y & = \frac{x^2+3}{x} \\
xy & =x^2+3 \\
x^2-xy+3&=0 
\end{align*}

Solve for x using the quadratic formula. We have a=1,\:b=-y,\:\text{and}\:c=3

\begin{align*}
x & =\frac{-b\pm \sqrt{b^2-4ac}\:}{2a} \\
x & =\frac{ -\left(-y\right)\pm \sqrt{\left(-y\right)^2-4\left(1\right)\left(3\right)}}{2\left(1\right)} \\
x & =\frac{y\pm \sqrt{y^2-12}}{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 1

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PROBLEM:

If \displaystyle f\left(x\right)=x^2-4x, find

a) \displaystyle f\left(-5\right)

b) \displaystyle f\left(y^2+1\right)

c) \displaystyle f\left(x+\Delta x\right)

d) \displaystyle f\left(x+1\right)-f\left(x-1\right)

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SOLUTION:

Part A

\begin{align*}
f\left(-5\right) & =\left(-5\right)^2-4\left(-5\right)\\
& =25+20\\
& =45 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

\begin{align*}
f\left(y^2+1\right) & = \left(y^2+1\right)^2-4\left(y^2+1\right)\\
& =y^4+2y^2+1-4y^2-4\\
& =y^4-2y^2-3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

\begin{align*}
f\left(x+\Delta x\right)&=\left(x+\Delta x\right)^2-4\left(x+\Delta x\right)\\
& =\left(x+\Delta x\right)\left[\left(x+\Delta x\right)-4\right]\\
& =\left(x+\Delta x\right)\left(x+\Delta x-4\right) \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ 
\end{align*}

Part D

\begin{align*}
f\left(x+1\right)-f\left(x-1\right) & =\left[\left(x+1\right)^2-4\left(x+1\right)\right]-\left[\left(x-1\right)^2-4\left(x-1\right)\right]\\
& = \left[x^2+2x+1-4x-4\right]-\left[x^2-2x+1-4x+4\right]\\
& =x^2-x^2+2x-4x+2x+4x+1-4-1-4\\
& =4x-8\\
& =4\left(x-2\right) \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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