Author Archives: Engineering Math

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

Advertisements
Advertisements

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).


Advertisements
Advertisements

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*}

Advertisements
Advertisements

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

Advertisements
Advertisements

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.


Advertisements
Advertisements

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*}

Advertisements
Advertisements

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

Advertisements
Advertisements

PROBLEM:

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


Advertisements
Advertisements

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*}

Advertisements
Advertisements

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

Advertisements
Advertisements

PROBLEM:

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


Advertisements
Advertisements

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*}

Advertisements
Advertisements

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

Advertisements
Advertisements

PROBLEM:

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


Advertisements
Advertisements

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*}

Advertisements
Advertisements

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

Advertisements
Advertisements

PROBLEM:

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


Advertisements
Advertisements

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*}

Advertisements
Advertisements

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1 Problem 1

Advertisements
Advertisements

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)


Advertisements
Advertisements

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*}

Advertisements
Advertisements

College Physics by Openstax Chapter 2 Problem 31


A swan on a lake gets airborne by flapping its wings and running on top of the water.

(a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates  from rest at an average rate of  0.350 m/s2, how far will it travel before becoming airborne?

(b) How long does this take?


Solution:

Part A

We are given the following values: v_f=6.00\:\text{m/s}; v_0=0\:\text{m/s}; and a=0.350\:\text{m/s}^2.

From the kinematic equations, the most applicable formula to solve for the change in distance, \Delta \text{x}, is

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

Solving for \Delta \text{x} in terms of the other variables, we have

\Delta x=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a}

Substituting the given values,

\begin{align*}
\Delta x & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a} \\
\Delta x  & =\frac{\left(6.00\:\text{m/s}\right)^2-\left(0.00\:\text{m/s}\right)^2}{2\left(0.350\:\text{m/s}^2\right)} \\
\Delta x  & =51.4286\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

From the formula v_f=v_0+at, solve for time, t in terms of the other variables.

t=\frac{v_f-v_0}{a}

Substitute the given values

\begin{align*}
t & =\frac{v_f-v_0}{a} \\
t & =\frac{6.00\:\text{m/s}-0.00\:\text{m/s}}{0.350\:\text{m/s}^2} \\
t & =17.1429\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Advertisements
Advertisements

College Physics by Openstax Chapter 2 Problem 30


A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of 0.250 m.

(a) How long did the acceleration last?

(b) Calculate the acceleration.


Solution:

We are given the following values:v_0=0\:\text{m/s}; v_f=65.0\:\text{m/s}; and \Delta x=0.250\:\text{m}.

We can immediately solve for the acceleration using the given values, so we are going to answer Part B first.

Part B

Solve for the acceleration first using the formula

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

We solve for acceleration in terms of the other variables.

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substitute the given values

\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & = \frac{\left(65.0\:\text{m/s}\right)^2-\left(0\:\text{m/s}\right)^2}{2\left(0.250\:\text{m}\right)} \\
a & =8450\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part A

To solve for the time of this motion, we shall use the formula

v_f=v_0+at

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

t=\frac{v_f-v_0}{a}

We now substitute the values given, and the computed acceleration to find the time.

\begin{align*}
t & =\frac{v_f-v_0}{a} \\
t & =\frac{65.0\:\text{m/s}-0\:\text{m/s}}{8450\:\text{m/s}^2} \\
t & =7.6922\:\times 10^{-3}\:\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Advertisements
Advertisements