# Tutor profile: Madeleine W.

## Questions

### Subject: SAT

A population of wild thrushes is observed between the years 1950 and 2000. On January 1, 1950, the wild thrush population had a size of $$1,389$$. According to the scientists’ data, the population size of the wild thrushes increased by $$36$$% every $$5$$ years. What equation correctly models the population size $$P$$ in terms of number of years $$n$$? A) $$P(n)=1,389(0.36)^{5n}$$ B) $$P(n)=1,389(1.36)^{\frac{n}{5}}$$ C) $$P(n)=50(1.36)^{5n}$$ D) $$P(n)=1,389(1.36)^{5n}$$

The correct answer is B, $$P(n)=1,389(1.36)^{\frac{n}{5}}$$. The starting population size is $$1,389$$, which immediately rules out answer C. We are told that the wild thrush population increases by $$36$$%. In the model, the starting population must therefore be multiplied by $$1.36$$, since multiplying the starting population by $$0.36$$ would lead to a decrease in the population. Thus answer A may also be ruled out. All we need to do now is choose the correct exponent. If the population size increases by $$36$$% every $$5$$ years, we can expect $$10$$ ‘checkpoints’ of $$1,389$$ being increased by $$36$$% between 1950 and 2000, a $$50$$-year time span. If we plug in $$50$$ for $$n$$ in answer D, $$P(n) = 1,389(1.36)^{5n}$$, we get: $$P(50)=1,389(1.36)^{5 \times 50}=1,389(1.36)^{250}$$. We see that the original population size will be multiplied $$250$$ times by $$1.36$$—in other words, the population size of the wild thrushes will increase by $$36$$% $$250$$ times over the course of 50 years. That is clearly not what is depicted in the problem, thus ruling out answer D. If we plug in $$50$$ for $$n$$ in answer B, $$P(n)=1,389(1.36)^{\frac{n}{5}}$$, we get: $$P(n)=1,389(1.36)^{\frac{50}{5}}=1,389(1.36)^{10}$$. We see that the original population size increases by $$36$$% $$10$$ times over the course of 50 years, which is exactly what is depicted in the problem. Thus B is the correct answer. $$\square$$

### Subject: Calculus

A student in a woodworking class would like to make a math-inspired masterpiece. Here is her process: 1. She chooses $$f(x)=\sin(x)$$ as her inspirational function. 2. She elects to make the maximum ‘height’ of her sculpture (the amplitude of the sine wave) $$1$$ ft, and she decides to make the ‘length’ of her sculpture (half the period of the sine wave) $$\pi$$ ft. 3. She creates her sculpture by imagining that $$f(x)=\sin(x)$$ is rotated $$180$$ degrees about the $$x$$-axis into the plane of vision. The final sculpture looks like a hill which, when looked at from the side, has the shape of $$f(x)=\sin(x)$$, and which, when cut perpendicular to the $$x$$-axis, has the shape of a semi-circle. What is the volume of the student’s final sculpture?

When learning about integration, we are usually asked to imagine the area ‘beneath a curve’ as the sum of the areas of an infinite amount of rectangles, each with an infinitesimal thickness. Likewise, we can find the volume of the final sculpture in the problem by imagining the sculpture as being divided into an infinite amount of three-dimensional slivers of wood, each with an infinitesimal ‘thickness’ or ‘depth.’ We will find the volume of the entire sculpture by adding up, through integration, the volumes of all these slivers of wood. There are a few ways to ‘cut’ the sculpture—a sine wave rotated $$180$$ degrees about the $$x$$-axis into the plane of vision—but the most helpful way to cut this sculpture is by cutting perpendicular to the $$x$$-axis (along the $$y$$-axis). When we do so, the individual ‘slices’ of the sculpture will look like semi-circles with infinitesimal depth. We next need to find the volume of each three-dimensional semi-circle. — Each 3D semi-circle will have a depth that can be represented as $$\Delta x$$. — The radius of the 3D semi-circle will depend on the ‘height’ of the sculpture at each placement of the ‘slice.’ Since the side-view of the sculpture is defined by the function $$f(x)=\sin(x)$$, the ‘height’ of each 3D semi-circle will in fact be $$\sin(x)$$. — The volume of the 3D semi-circle will be the area of each semi-circle times $$\Delta x$$: $$\frac{1}{2} \pi r^2 \Delta x= \frac{1}{2} \pi (\sin(x))^2 \Delta x$$. Now that we have the volume of each 3D semi-circle, we can add up the volumes using integration. Since the ‘length’ of the sculpture is a half of a period of a regular sine wave, we know that we must integrate the volumes of the slices from $$0$$ to $$\pi$$: $$V = \int_{0}^{\pi} \frac{1}{2} \pi \sin^2(x) dx$$. Moving the constants outside of the integral, we have: $$\frac{1}{2} \pi \int_{0}^{\pi} \sin^2(x)dx$$. To find the integral of $$\sin^2(x)$$, we use the trigonometric identity $$\sin^2(x)=\frac{1-\cos(2x)}{2}$$: $$\int_{0}^{\pi}\sin^2(x)dx = \int_{0}^{\pi} \frac{1-\cos(2x)}{2}dx = \int_{0}^{\pi} \frac{1}{2}(1-\cos(2x))dx$$. Moving the constant outside of the integral, we have: $$\frac{1}{2}\int_{0}^{\pi}(1-\cos(2x))dx$$. Applying the Sum Rule, we have: $$\frac{1}{2}(\int_{0}^{\pi}1dx - \int_{0}^{\pi}\cos(2x)dx)$$. $$\int_{0}^{\pi}1dx = [x]_{0}^{\pi} = \pi - 0 = \pi$$. $$\int_{0}^{\pi}\cos(2x)dx = [\frac{1}{2}\sin(2x)]_{0}^{\pi} = \frac{1}{2}\sin(2\pi)-\frac{1}{2}\sin(0) = 0$$. $$\Rightarrow \int_{0}^{\pi}\sin^2(x)dx=\frac{1}{2}(\pi-0)=\frac{\pi}{2}$$. The volume of the sculpture is therefore: $$\frac{1}{2}\pi\int_{0}^{\pi}\sin^2(x)dx=\frac{\pi}{2} \times \frac{\pi}{2} = \frac{\pi^2}{4}$$ ft$$^3$$. $$\square$$

### Subject: Algebra

A large water bottle in the shape of a perfect cylinder is being filled at a rate of $$3 \frac{L}{min}$$, but has sprung a leak that allows water to flow out of the bottle at a rate of $$1 \frac{L}{min}$$. a) Write the height $$h$$ of the cylindrical bottle in terms of volume $$V$$ and radius $$r$$. b) If the volume of the water bottle is 3 liters, and the diameter of the water bottle is 4 inches, what is the height of the water in the water bottle after 1 minute? Assume that $$1 L = 61 in^3$$, and use $$\pi = 3.14$$. Round your number to the nearest hundredth inch.

a) The volume of a cylinder is $$V = \pi r^2 h$$. To write $$h$$ in terms of $$V$$ and $$r$$, divide both sides of the by $$\pi r^2$$: $$\frac{V}{\pi r^2} = \frac{\pi r^2 h}{\pi r^2} \Rightarrow h = \frac{V}{\pi r^2}$$ b) We first must figure out the volume of water in the water bottle at the 1-minute mark. Since water is flowing into the water bottle at the rate of $$3 \frac{L}{min}$$, but flowing out of the water bottle at the rate of $$1 \frac{L}{min}$$, the volume of water in the water bottle after 1 minute will be: $$3 L - 1 L = 2 L$$. It is helpful to next convert the volume from liters to inches cubed, since the question requires an answer in inches. We do this by using a simple unit conversion: $$2L \times \frac{61 in^3}{1L} = (2 \times 61) in^3 = 122 in^3$$. Also, we note that the radius of the cylinder is $$\frac{4}{2} = 2$$ inches. Plugging this information into our original equation for $$h$$, and using the given $$\pi = 3.14$$, we have: $$h = \frac{122}{3.14 \times 2^2}=\frac{122}{3.14 \times 4} = \frac{122}{12.56} = 9.71$$ in. $$\square$$

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