Author Archives: Engineering Math

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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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: vf=6.00 m/sv_f=6.00\:\text{m/s}; v0=0 m/sv_0=0\:\text{m/s}; and a=0.350 m/s2a=0.350\:\text{m/s}^2.

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

(vf)2=(v0)2+2aΔx\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

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

Δx=(vf)2−(v0)22a\Delta x=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a}

Substituting the given values,

Δx=(vf)2−(v0)22aΔx=(6.00 m/s)2−(0.00 m/s)22(0.350 m/s2)Δx=51.4286 m  (Answer)\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 vf=v0+atv_f=v_0+at, solve for time, tt in terms of the other variables.

t=vf−v0at=\frac{v_f-v_0}{a}

Substitute the given values

t=vf−v0at=6.00 m/s−0.00 m/s0.350 m/s2t=17.1429 s  (Answer)\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*}
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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:v0=0 m/sv_0=0\:\text{m/s}; vf=65.0 m/sv_f=65.0\:\text{m/s}; and Δx=0.250 m\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

(vf)2=(v0)2+2aΔx\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

We solve for acceleration in terms of the other variables.

a=(vf)2−(v0)22Δxa=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substitute the given values

a=(vf)2−(v0)22Δxa=(65.0 m/s)2−(0 m/s)22(0.250 m)a=8450 m/s2  (Answer)\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

vf=v0+atv_f=v_0+at

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

t=vf−v0at=\frac{v_f-v_0}{a}

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

t=vf−v0at=65.0 m/s−0 m/s8450 m/s2t=7.6922 ×10−3 s  (Answer)\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*}
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Mortgage Project

You can buy the solution to this project for only $10 right here.

A home will likely be the biggest purchase a person ever makes, along with being the most intimidating purchase. But the suitable news is most of the problems homebuyers face have a quick solution, if accomplished before trying to get a mortgage.

Also the purpose of this project is help you to overcome the imitating home buying process, definitely it will gave a chance to get familiar with the process:

  • Check your credit report so you are mindful of what your current credit score is before for a loan. Credit reporting agencies must give you one free report annually.
  • Try to work closely with your banker to figure out how much you can borrow and which loan exactly fits you.
  • Acquire what current mortgage rates are. Bankers are there to help you understand how that translates into monthly mortgage payments and that’s the purpose of this project.

The following formula is used for figuring out a monthly home mortgage payment:

\displaystyle M=\frac{Lr\left[1+\frac{r}{12}\right]^{12t}}{12\left[\left(1+\frac{r}{12}\right)^{12t}-1\right]}

where:           L=the loan amount in dollars
                        r = the annual interest rate expressed as decimal
                        t = the number of years of the loan
                       M = the monthly payment in dollars

You are looking to buy a $325,000 home in Haverhill. If Bank of America will give them a 30-year mortgage at 6% annual interest rate for the cost of the house after they receive a 10% down payment.

A. Determine the loan amount?

B. How much their monthly payment will be?  

C. At the end of the 30-years, how much total money will you have paid to Bank of America for your home? In another word how much did the $325,000 house really cost the couple? 

D. How much interest will they have paid?       

E. How many of her monthly payment go toward the interest?

F. What percent increase over the cost of the home does this interest represent? 

G. Redo and re-answer all questions, but this time for 15 years?

Do analysis comparison between 30 and 15 years mortgage (at least one page not double spacing).

 

College Physics by Openstax Chapter 2 Problem 29

Freight trains can produce only relatively small accelerations and decelerations.

(a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 minutes, starting with an initial velocity of 4.00 m/s?

(b) If the train can slow down at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity?

(c) How far will it travel in each case?

Solution:

Part A

We are given the the following: a=0.0500 m/s2a=0.0500 \ \text{m/s}^2; t=8.00 minst=8.00 \ \text{mins}; and v0=4.00 m/sv_0=4.00 \ \text{m/s}.

The final velocity can be solved using the formula vf=v0+atv_f=v_0+at. We substitute the given values.

vf=v0+atvf=4.00 m/s+(0.0500 m/s2)(8.00 min×60 sec⁡1 min⁡)vf=28.0 m/s  (Answer)\begin{align*} v_f& = v_0+at \\ v_f & = 4.00\:\text{m/s}+\left(0.0500\:\text{m/s}^2\right)\left(8.00\:\text{min}\times \frac{60\:\sec }{1\:\min }\right) \\ v_f & = 28.0 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Rearrange the equation we used in part (a) by solving in terms of tt, we have

t=vf−v0at=0 m/s−28 m/s−0.550 m/s2t=50.91 sec⁡  (Answer)\begin{align*} t & =\frac{{v_f}-v_0}{a} \\ t & = \frac{0\:\text{m/s}-28\:\text{m/s}}{-0.550\:\text{m/s}^2} \\ t & = 50.91\:\sec\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The change in position for part (a), Δx \Delta x, or distance traveled is computed using the formula Δx=v0t+12at2 \Delta x=v_0 t+\frac{1}{2} at^2.

Δx=v0t+12at2Δx=(4.0 m/s)(480 s)+12(0.0500 m/s2)(480 s)2Δx=7680 m  (Answer)\begin{align*} \Delta x & =v_0 t+\frac{1}{2} at^2 \\ \Delta x & =\left(4.0\:\text{m/s}\right)\left(480\:\text{s}\right)+\frac{1}{2}\left(0.0500\:\text{m/s}^2\right)\left(480\:\text{s}\right)^2 \\ \Delta x & = 7680\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

For the situation in part (b), the distance traveled is computed using the formula Δx=vf2−v022a\Delta x=\frac{v_f^2-v_0^2}{2 a}.

Δx=(0 m/s)2−(28.0 m/s)22(−0.550 m/s2)Δx=712.73 m  (Answer)\begin{align*} \Delta x & =\frac{\left(0\:\text{m/s}\right)^2-\left(28.0\:\text{m/s}\right)^2}{2\left(-0.550\:\text{m/s}^2\right)} \\ \Delta x & =712.73\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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