Tag Archives: 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