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


A right circular cylinder, 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).


Solution:

Refer to the figure below.

1.1.7

By ratio and proportion of two similar triangles ΔBCD and ΔACE, we have

\displaystyle \frac{y}{r-x}=\frac{h}{r}

\displaystyle y=\frac{h\left(r-x\right)\:}{r}


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


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.


Solution:

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

\displaystyle S=bd^{3\:}

\displaystyle S=20d^3


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


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


Solution:

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

\displaystyle A=\frac{1}{2}\cdot x\cdot x\cdot sin\:60^{\circ}

\displaystyle A=\frac{1}{2}\cdot x^2\cdot \frac{\sqrt{3}}{2}

\displaystyle A=\frac{\sqrt{3}}{4}x^2


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


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


Solution:

\displaystyle Distance=Rate\times Time

\displaystyle D=\left(60\:km/hr\right)\cdot t

\displaystyle D=60t


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


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


Solution:

\displaystyle y=tan\left(x+\pi \right)

\displaystyle x+\pi =tan^{-1}y

\displaystyle x=tan^{-1}y-\pi


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


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


Solution:

\displaystyle y=\frac{x^2+3}{x}

\displaystyle xy=x^2+3

\displaystyle x^2-xy+3=0

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

\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}\:}{2a}

\displaystyle x=\frac{\cdot -\left(-y\right)\pm \sqrt{\left(-y\right)^2-4\left(1\right)\left(3\right)}}{2\left(1\right)}

\displaystyle x=\frac{y\pm \sqrt{y^2-12}}{2}


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