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Katherine H.
High school math tutor for three years, current student at Princeton University
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Mandarin
TutorMe
Question:

请为下面的电子邮件写一封回信。 发件人:国华 邮件主题:欢迎新生 我和几个同学正在讨论下个学期怎么欢迎从国外来的学生。我们没有什么钱,请你帮我们想一个不用花很多钱的欢迎活动吧。另外,你觉得是开学前一两天还是开学那一天办活动比较好?为什么呢?先谢啦!

Katherine H.
Answer:

国华你好! 我建议你们可以去学校附近的一家中餐馆一起吃饭。中餐馆的菜的价格一般不是很贵,而且国外的学生还能品尝到中国的美食,有一个新的体验。我觉得开学前一两天办活动比较好,因为开学当天同学们都很忙,大家又要上课,很难找到一个合适的时间办活动。开学前一两天大家的时间都比较充裕。 祝你们办活动成功! 小明

Geometry
TutorMe
Question:

Semicircles with centers at $$A$$ and $$B$$ and with radii $$2$$ and $$1$$, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $$\overline{JK}$$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at $$P$$ is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at $$P$$?

Katherine H.
Answer:

Connect $$AP$$, $$BP$$, and $$OP$$. Using laws of cosines on $$\triangle BOP$$ with $$\angle B$$ as the reference angle, we obtain $$ (3-r)^2 = (1+r)^2 + 2^2 - 2 \times 2(1+r) \cos B$$. Solving for $$\cos B$$, we get $$\cos B = \frac{2r-1}{r+1}$$. Using laws of cosines on $$\triangle BPA$$ with $$\angle B$$ as the reference angle, we obtain $$ (2+r)^2 = (1+r)^2 + 3^2 - 2 \times 3(1+r) \cos B$$. Solving for $$\cos B$$, we get $$\cos B = \frac{3-r}{3(r+1})$$. Equate the two expressions of $$\cos B$$, we find $$r = \frac{6}{7}$$.

Calculus
TutorMe
Question:

Calculate $$I = \int_{-\infty}^{\infty} e^{-x^2} dx$$

Katherine H.
Answer:

In this problem, we use a double integral to calculate the single integral. Because $$e^{-x^2}$$ is an even function, we can take $$\frac{I}{2} = \int_{0}^{\infty} e^{-x^2} dx$$. Since x is just a dummy variable, $$\int_{0}^{\infty} e^{-x^2} dx = \int_{0}^{\infty} e^{-y^2} dy$$. So we can multiply the two integral of same value but different dummy variables: $$\frac{I}{2} \cdot \frac{I}{2} = \int_{0}^{\infty} e^{-x^2} dx \cdot \int_{0}^{\infty} e^{-y^2} dy = \int_{0}^{\infty} \int_{0}^{\infty} e^{-(x^2+y^2)} dxdy$$. The form $$x^2 + y^2$$ inspires us to change variables x and y into polar coordinates: $$ x = r\cos(\theta), y = r\sin(\theta)$$. After change of variables, we obtain $$ \frac{I^2}{4} = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} re^{-r^2}drd\theta = \int_{0}^{\frac{\pi}{2}} -\frac{1}{2} e^{-r^2} |_{0}^{\infty}d\theta = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} d\theta = \frac{\pi}{4}$$.

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