Category Archives: Calculus

Includes differential, integral and series calculus

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

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PROBLEM:

If f(x)=3x2−4x+1\displaystyle f\left(x\right)=3x^2-4x+1, find f(h+3)−f(3)h, h≠0\displaystyle \frac{f\left(h+3\right)-f\left(3\right)}{h},\:h\ne 0.

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SOLUTION:

f(h+3)−f(3)h=[3(h+3)2−4(h+3)+1]−[3(3)2−4(3)+1]h=3(h2+6h+9)−4h−12+1−16h=3h2+18h+27−4h−12+1−16h=3h2+14hh=h(3h+14)h=3h+14  (Answer)\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 f(x)=x2+1\displaystyle f\left(x\right)=x^2+1, find f(x+h)−f(x)h, h≠0\displaystyle \frac{f\left(x+h\right)-f\left(x\right)}{h},\:h\ne 0.

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SOLUTION:

f(x+h)−f(x)h=[(x+h)2+1]−(x2+1) h=x2+2xh+h2+1−x2−1h=2xh+h2h=h(2x+h)h=2x+h  (Answer)\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 xx, height yy, is inscribed in a right circular cone, radius of base rr and a height hh. Express yy as a function of xx (rr and hh 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

yr−x=hry=h(r−x) r  (Answer)\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 SS=stiffness, bb=breadth, and dd=depth

S=bd3S=20d3  (Answer)\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 AA of an equilateral triangle as a function of its side xx.

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SOLUTION:

From the formula of the area of a triangle, A=12absin⁡(θ)\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 sin⁡ 60∘=32\displaystyle \sin\:60^{\circ} =\frac{\sqrt{3}}{2}.

A=12absin⁡(θ)A=12⋅x⋅x⋅sin⁡ 60∘A=12⋅x2⋅32A=34x2  (Answer)\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 DD traveled in tt hr by a car whose speed is 60 km/hr.

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SOLUTION:

Distance=Rate×TimeD=(60 km/hr)⋅t hrD=60t km  (Answer)\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 y=tan⁡(x+π) \displaystyle y= \tan\left(x+\pi \right), find xx as a function of yy.

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SOLUTION:

y=tan⁡(x+π)x+π=tan⁡−1yx=tan⁡−1y−π  (Answer)\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 y=x2+3x\displaystyle y=\frac{x^2+3}{x}, find xx as a function of yy.

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SOLUTION:

y=x2+3xxy=x2+3x2−xy+3=0\begin{align*} y & = \frac{x^2+3}{x} \\ xy & =x^2+3 \\ x^2-xy+3&=0 \end{align*}

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

x=−b±b2−4ac 2ax=−(−y)±(−y)2−4(1)(3)2(1)x=y±y2−122  (Answer)\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 f(x)=x2−4x\displaystyle f\left(x\right)=x^2-4x, find

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

b) f(y2+1)\displaystyle f\left(y^2+1\right)

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

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

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SOLUTION:

Part A

f(−5)=(−5)2−4(−5)=25+20=45  (Answer)\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

f(y2+1)=(y2+1)2−4(y2+1)=y4+2y2+1−4y2−4=y4−2y2−3  (Answer)\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

f(x+Δx)=(x+Δx)2−4(x+Δx)=(x+Δx)[(x+Δx)−4]=(x+Δx)(x+Δx−4)  (Answer)\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

f(x+1)−f(x−1)=[(x+1)2−4(x+1)]−[(x−1)2−4(x−1)]=[x2+2x+1−4x−4]−[x2−2x+1−4x+4]=x2−x2+2x−4x+2x+4x+1−4−1−4=4x−8=4(x−2)  (Answer)\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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