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Meaghan L.
High School Math Teacher for 8 Years
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Pre-Calculus
TutorMe
Question:

A Ferris wheel whose maximum height reaches 100 feet sits 7 feet above the ground. This wheel makes 2 rotations per minute. Model a person’s height (feet) on the Ferris wheel over time (seconds) using a sinusoidal function.

Meaghan L.
Answer:

Sine and Cosine graphs can be very helpful when trying to model something that happens in cycles. A Ferris wheel for example goes up and down. So if we are modeling height over time our graph would wave up and down. Cosine works perfect for this particular function because on a Ferris wheel you state at the lowest point on the wheel. Likewise y=–cos(x) starts at a minimum. We will need to change the following: Mid line (the half-way point between the max and minimum height). This will be represented as k Amplitude (how high and how low the graph travels from the mid-line) This will be represented as a Period (how long it take the graph to complete on rotation) This will be represented as 360/b First let’s remember where those numbers go in the equation. y=-a cos⁡(bx)+k Mid line: If you are riding this particular Ferris wheel how high off the ground are you when you are half way to the top of the wheel? The max height is 100 feet, but the wheel is 7 feet of the ground. 100-7=93. So the diameter of the wheel is 93 feet. The radius is half of that 93/2=46.5. So you have the radius of 46.5 plus the 7 feet from the bottom of the wheel to the ground. The answer to the question, “How high off the ground are you when you are half way to the top of the wheel?” is then 53.5 ft. This is the mid-line. Adding 53.5 to the end of our equation will bring that “halfway” mark to 53.5 ft. Therefore k=53.5. Amplitude: What is the difference between your max and minimum heights? Half of that is the amplitude. In other words, the amplitude is the distance from the mid-line to the max/min. Wouldn’t that just be the radius of our wheel? Of course! 46.5 ft. If we multiply our cosine function by 46.5 it will stretch the max and min to 7 and 100. Therefore a=46.5 Period: Now that we have fixed the amplitude and mid-line we need to make sure that 2 rotations takes 1 minute. As the equation is now one rotation takes 360 minutes. This is because the period of cosine if not changes is 360°. The question is if 2 rotations takes 1 minute, how long does 1 rotation take? The answer is 30 seconds. So we need to take the period from 360 to 30. We can do this by multiplying the x-value of our function by 12. Therefore b=12. If you plug in a, k, and b into the cosine function you get. y=-46.5cos⁡(12x)+53.5. Check your answer by plugging in 30 for x. You should get 7 since at 30 seconds you would have made it all the way back to the lowest point of the wheel.

Calculus
TutorMe
Question:

A 6ft tall man is walking away from a 20ft lamppost at a rate of 5 ft/min. How fast is his shadow growing when he is 9 feet from the poll?

Meaghan L.
Answer:

Let x be the distance the man is from the light pole. Remember that this is a changing value. Let s be the length of the shadow. This is also changing. The rate at which it is changing is what we are trying to find. The height of the lamppost and the man’s height are fixed values. We are given the rate at which he is walking away (5ft/min). This means that x is changing at a rate of 5ft/min. So we can say that dx/dt is 5. We are looking for ds/dt. We need a formula to relate these rates (hence, related rates problems). To do these we will use some properties of similar triangles. There are two similar triangles in our picture. The big one with legs 20 and (x+s) and the smaller one with legs 6 and s. We can set up a proportion with the side lengths since “corresponding parts of congruent triangles are proportional.” That will look like this: 20/(x+s)=6/s We will use implicit differentiation so we can plug in the information we now. Let’s cross multiply and simplify to make that easier. 20s=6(x+s) 20s=6x+6s 14s=6x Use implicit differentiation to find this derivative. This gives us: 14(ds/dt)=6(dx/dt) Plug in what we are given (dx/dt) and solve for (ds/dt). 14(ds/dt)=6*5 14(ds/dt)=30 ds/dt=30/14=2.14 ft/min

Algebra
TutorMe
Question:

Write a linear equation in slope intercept form to model the following situation, then answer the question that follow: Its Saturday morning and you have decided to ride your bike to your friends house to spend the day. She lives 7 miles away from you. You usually average 15 mph when riding your bike. 1. Write an equation to model your distance from your friends house (y) in terms of the about of time (in hours) you have been riding (x).

Meaghan L.
Answer:

When we are writing an equation in slope intercept form (y=mx+b) we need to know two things, the slope of the line (m) and the y-intercept (b). Here we are given two variables: Your distance from your friends house (y) and the amount of time you have been riding (x). Lets start with the slope. This value will be substituted in our equation for m. Slope is the rate your y is changing compared to your x. In other words, how fast is your distance changing as your ride your bike for one hour? 15 miles! Since the distance is decreasing as you ride we will say the slope (m) is -15. Now for the y-intercept (b). This is our starting point. What is the distance before you start riding? 7 miles! Because your friend lives 7 miles away. Our answer is then y=-15x+7. You can plug in some numbers for x to make sure you get a distance that makes sense. If you are looking for a challenge, use the equation to see how long it will take you to get there. What important point would this value be on the graph of y=-15x+y?

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