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# Tutor profile: Yana N.

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Yana N.
Mathematics and Statistics Tutor
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## Questions

### Subject:Trigonometry

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Question:

Rachel takes her little sister to an amusement park. There are two Ferris wheels; a small one for children under 12 and a much larger one. The small wheel reaches 8 meters at its highest point, and the cabin is 0.6 meters from the ground at its lowest point. The small wheel makes 3 revolutions in 2 minutes. The bigger wheel reaches a height of 25 meters, with the cabins hanging 1.5 meter from the ground. It makes one revolution per minute. Rachel is going to ride the big wheel, while her sister rides the small one. Since there are trees between the two wheels, They can only see each other on the rides when Rachel is more than 5 meters, but less than 10 meters, in the air. Her sister must also be at least 5 meters in the air. Create a model of the two Ferris wheels and determine when Rachel and her sister can see each other during the first minute of their rides.

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Yana N.

First we must come up with a model for each of the Ferris wheels. Consider the following equation: $$y=A cos(Bx+C)+D$$ where $$A, B, C,$$ and $$D$$ are constants. $$A$$ will determine the amplitude of the cosine wave, $$B$$ determines the period, and $$C$$ and $$D$$ determine the shift across the $$x$$ and $$y$$ axes respectively. Lets start by creating a model for the smaller wheel. If we start with $$H(t)=-cos(t)$$, This gives us the right shape for modelling the height of someone riding a Ferris wheel. To work out $$A$$, we must consider the height of the Ferris wheel. The basic $$cos(x)$$ function where $$A=1$$ oscillates between -1 and 1 ($$A$$ is the half the distance between the lowest and highest point). Since the Ferris wheel starts at a height of 0.6 meters and reaches a height of 8 meters, $$A=3.7$$ since the total diameter of the wheel is 7.4 meters. Now we must find $$B$$, the period. The basic $$cos(x)$$ where $$B=1$$ has a period of $$2\pi$$. However, we need to alter our period to correspond to the the time it takes for the Ferris wheel to make a revolution. Since the wheel makes 3 revolutions in 2 minutes, so we know that the wheel will make one full revolution every 40 seconds. So, to get $$B$$ for our equation, all we have to do is divide our original period ($$2\pi$$) by our new period (40 seconds) to get $$B=\frac{2\pi}{40}=\frac{\pi}{20}$$ Since the negative cosine function already provides the right shape, we can stick with the original $$C$$ value, which is 0. Lastly, we must determine the vertical shift of the model. Currently our model is centered around zero. We want to bring up the entire function so that at any given point, $$H(t)$$ is positive, we also need to consider the upper and lower bounds, where the function must never go below 0.6 meters (as that is the lowest point that the Ferris wheel reaches) and never above 8 meters (the highest point). As such, we need to center our function between these values and we get $$D=4.3$$, which would be the center of the Ferris wheel. Now we can sub in these values to get $$H(t)=-3.7cos(\frac{\pi }{20}t) +4.3$$ as the model for the height of the cabin on the Ferris wheel over time for the small wheel. We can repeat this process to get $$H(t)=-(\frac{25-1.5}{2} )cos(\frac{2\pi}{60} t+0) +13.25 =-11.75cos(\frac{\pi }{30}t)+13.25$$ as our equation for the big Ferris wheel. Now we must find out when both Ferris wheels are 5 meters above the ground, so we set $$H(t)=5$$ and solve as follows $$5=-3.7cos(\frac{\pi t}{20}) +4.3$$ $$0.7=-3.7cos(\frac{\pi t}{20})$$ (Subtract 4.3 from both sides) $$\frac{-7}{37}=cos(\frac{\pi t}{20})$$ (Divide both sides by -3.7) $$cos^{-1}(\frac{-7}{37})=\frac{\pi t}{20}$$ (Take the inverse cosine of both sides) $$\frac{20cos^{-1}(\frac{-7}{37})}{\pi}=t$$ (Divide both sides $$\frac{\pi}{20}$$) and we get $$t_{1}=11.21171889...$$ and since the curve is symmetric, we can find the second solution where $$H(t)=5$$ in the first oscillation by subtracting this value from the period to get $$t_{2}=28.78828111....$$ and we can get the time where she is 5 meter above ground during the second revolution by adding 40 seconds to each of those times to get that in the first minute of being on the ride (the time it takes the $$big$$ to complete a revolution), Rachel's sister is more than 5 meters above the ground 11.2 seconds after getting onto the ride until 28.8 seconds after starting the ride where she dips below 5 meters again, and once more at 51.2 seconds until 68.8 We do the same to find out that Rachel is 5 meters above the ground between $$t_3=7.567007223...$$ and $$t_4=52.43299278...$$. But Rachel 10 meters or more above the ground between $$t_5=12.32381121...$$ and $$t_6=47.67618879...$$. So Rachel is only more than 5 but less than 10 meters above the ground from 7.6 seconds into the ride until 12.3 seconds into the ride and from 47.7 seconds into the ride until 52.4 seconds. Looking at the times they are both in the right height range, they can only briefly see each other for about a second at 11.2-12.3 seconds into the ride and and 51.2-52.4 seconds into the ride.

### Subject:Calculus

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Question:

Volumes of Revolution - Shell method $$V = \int_{a}^{b}2 \pi xy$$ $$dy$$ You are manufacturing an ellipsoid bead for a jeweler. He has specified that the length of the bead must be no longer than 15mm and the width 5mm, The bead needs to have a hole 1mm wide down the length so that it can be threaded. He has provided you with 1m$$^3$$ of material. How many beads can you make?

Inactive
Yana N.

In order to find the how many beads we can manufacture, we must first determine the volume of the bead he has requested. Using the volumes of revolution method, we must first model a cross-section of the bead, which can be approximated by the equation of an ellipse $$E=\frac{x^2}{length^2}+\frac{y^2}{width^2}=1$$ which gives us $$\frac{x^2}{225}+\frac{y^2}{25}=1$$ Since the volume is calculated on rotation, we only need to consider half of the cross-section. Now we must choose our limits. If we revolve around the x-axis, then the limits of the integral must be perpendicular, hence $$b=5$$ which would be the widest part of the bead and $$a= 0.5$$ since we must account for the threading hole. Now, in order to use our formula for Volume, we need to redefine $$x$$ in terms of $$y$$ as follows: $$x^2+9y^2=225$$ (Multiplying every term by 25) $$x^2=225-9y^2$$ (Subtract $$9y^2$$ from both sides) $$x^2=9(25-y^2)$$ (Factorize LHS) $$x=3\sqrt{25-y^2}$$ (Take the square root of both sides) $$x=3(25-y^2)^\frac{1}{2}$$ (Re-write the square root as a power) We can now substitute this into the formula: $$V=\int_{\frac{1}{2}}^{5} 2 \pi 3(25-y^2)^\frac{1}{2} y$$ $$dy$$ $$V=6 \pi \int_{\frac{1}{2}}^{5} (25-y^2)^\frac{1}{2} y$$ $$dy$$ and integrate. In order to integrate, we will employ the substitution method: Let $$u =25-y^2$$, then $$du = -2y$$ $$dy$$ or $$dy=\frac{-1}{2y} du$$. We must also redefine our bounds as follows, $$u=25-(5)^2=0$$ is the new upper bound and $$u=25-(\frac{1}{2})^2=\frac{99}{4}$$ is the new lower bound. Substituting this back into our original integral we get: $$V= 6 \int_{\frac{99}{4}}^{0} u^\frac{1}{2} y \frac{-1}{2y} du= -3\int_{\frac{99}{4}}^{0} u^\frac{1}{2} du$$ Integrating gives us $$V = -3 [ \frac{2}{3}u^\frac{3}{2}]_{\frac{99}{4}}^{0} = -2[u^\frac{3}{2}]_{\frac{99}{4}}^{0}$$ $$V= -2 [ (0)^\frac{3}{2} - (\frac{99}{4})^\frac{3}{2}] = 246.2593068...$$ So the volume of one bead is approximately 246.26mm$$^3$$ (2 dp) 1m$$^3$$ = 1000000mm$$^3$$ 1000000/246.2593068...=4060.7588495... You could make 4060 beads with the material provided.

### Subject:Algebra

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Question:

Simultaneous Equations Jennifer and Alice are organizing a fundraising event. They sell tickets for $15, but students can get a 20% discount. They have sold 95 tickets and have raised a total of$1335. Half of the tickets sold by Jennifer were sold at the discounted rate, while only one ninth of the the tickets sold by Alice were discounted. How many more tickets than Alice did Jennifer sell?

Inactive
Yana N.

First, we must organize the information we have; The cost of an regular ticket is $15, the cost of a student ticket is$12 Let $$x$$ be the number of regular tickets and $$y$$ be the number of student tickets sold. Since 95 tickets were sold in total, we can get equation $$(1)$$ $$x+y=95$$ From the total amount raised, and the cost of the tickets, we can get equation $$(2)$$ $$15x +12y = 1335$$ From here, we can rearrange the equation $$(1)$$ to get $$y=95-x$$ and substitute into the second equation to get $$15x +12(95-x) = 1335$$ If we simplify and solve: $$15x+1140-12x=1335$$ (Expanding the brackets) $$3x+1140=1335$$ (Combining like terms) $$3x=195$$ (Subtracting 1140 from both sides) $$x=65$$ (Dividing both sides by 3) Here, we get that the total number of regular tickets sold is 65, and therefore, by substituting 65 for $$x$$ in equation $$(1)$$, we get the number of student tickets to be 30. Let $$a$$ be the number of tickets sold by Jennifer, and $$b$$ be the number of tickets sold by Alice. From the initial brief, we can form another set of equations. $$(3)$$ $$a+b=95$$ (Since Jennifer and Alice sold all of the tickets) $$(4)$$ $$\frac{a}{2} + \frac{b}{9} = 30$$ (Where half of Jennnifer's tickets and $$\frac{1}{9}th$$ of Alice tickets make up the total number of student tickets sold) Following the same steps as the the first part, we rearrange equation $$(3)$$ to get $$a=95-b$$ and substitute into $$(4)$$ to get $$\frac{95-b}{2} + \frac{b}{9} = 30$$. Now we simplify again and solve as follows: $$(95-b) +\frac{2b}{9} = 60$$ (Multiply every term by 2) $$9(95-b)+2b = 540$$ (Multiply every term by 9) $$855-9b+2b=540$$ (Expanding the brackets) $$855-7b = 540$$ (Combining like terms) $$-7b=-315$$ (Subtracting 855 from both sides) $$b=45$$ (Dividing by -7) So, the total number of tickets sold by Alice was 45, and hence the total number of tickets sold by Jennifer is 50. Therefore, Jennifer sold 5 more tickets than Alice.

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