# Tutor profile: Rasimate M.

## Questions

### Subject: Pre-Calculus

Solve the equation $$(1.07)^t=2$$ for $$t$$.

The method to solve this problem is basic but useful in some topics in Calculus. First, we can take logarithms on both sides of the equation and get $(\ln(1.07)^t=\ln2.$)Then, using the logarithm property $$\log_ab^n=n\log_ab$$, we get$(t\ln(1.07)=\ln2.$)Finally, we use a calculator or WolframAlpha to approximate the value of $$t$$, $(t=\frac{\ln2}{\ln1.07}\approx10.24$) Note that to solve the equation, we can use other bases of logarithm such as 2 or 10.

### Subject: Pre-Algebra

Andy and Dan ordered two pizzas of the same size. Andy ate 3/4 of the first pizza and 1/3 of the second pizza. Dan ate the rest. Who ate more pizza?

From the problem, we can add the portion of the first pizza and the second pizza that Andy ate and get $(\frac{3}{4}+\frac{1}{3}=\frac{3\times3}{4\times3}+\frac{1\times4}{3\times4}=\frac{9}{12}+\frac{4}{12}=\frac{9+4}{12}=\frac{13}{12}=1\frac{1}{12}.$) So Andy ate more than one whole pizza. Since they ordered two pizzas, Andy ate more pizza than Dan. Note 1: We can also compute the portion of pizza that Dan ate and get $$\frac{11}{12}$$. So Dan ate less than one whole pizza and he ate less pizza than Andy. Note 2: We can draw figures as described in the problem and see that the one third of the second pizza that Andy ate is larger than the one forth of the first pizza that he did not eat. So the sum of portion of pizza that Andy ate is more that one whole pizza.

### Subject: Calculus

A vase is modeled by rotating the region bounded by the curve $$y=2+\sin(x)$$, the $$x$$-axis, the $$y$$-axis, and the line $$x=2\pi$$ about the $$x$$-axis. Neglecting the thickness of the vase, find its volume.

First, we sketch the described region and the vase obtained by rotating the region about the $$x$$-axis. From the sketch, we see that slicing the vase perpendicular to the $$x$$-axis through the point $$x$$ results in a circle. The radius of this circle is $$y=2+\sin(x)$$. These two observations suggest that we can use the cross-section method to find the volume. The area of this circle is $$\pi (2+\sin(x))^2$$. The cross-section method give the volume $(V=\int_0^{2\pi}\pi(2+\sin(x))^2dx=\pi\int_0^{2\pi}4+4\sin(x)+\sin^2(x)dx.$) To compute $$\int\sin^2(x)dx$$, we use the trigonometric identity $(\sin^2(x)=\frac{1-\cos(2x)}{2}.$) The integral can be calculated as follow. $(\int\sin^2(x)dx=\int\frac{1-\cos(2x)}{2}dx=\int\frac{1}{2}dx-\int\frac{\cos(2x)}{2}dx=\int\frac{1}{2}dx-\int\frac{\cos(2x)}{2}\frac{d2x}{2}=\frac{x}{2}-\frac{\sin(2x)}{4}+C.$) So the volume of the vase is $(V=\pi\int_0^{2\pi}4+4\sin(x)+\sin^2(x)dx=\pi[4x-4\cos(x)+\frac{x}{2}-\frac{\sin(2x)}{4}]_0^{2\pi}=\pi [(8\pi-4+\pi-0)-(0-4+0-0)]=9\pi^2 \text{ unit}^3.$)

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