# Tutor profile: Peter K.

## Questions

### Subject: Spanish

Translate the following sentence into Spanish: "I doubt that it is going to rain."

Since this sentence expresses doubt or uncertainty, we must use the present subjunctive tense. This is how you set it up: the first verb, "doubt" is conjugated normally (1st person present indicative), and it's followed by the word "que" and then the second verb, "going" (whose certainty is in question) is conjugated into the subjunctive. "To rain" is left in the infinitive form. So the sentence is translated into: "Dudo que vaya a llover."

### Subject: Calculus

Calculate the arclength of the curve determined by r = \sin\theta + \sin\theta from 0 to 2\pi.

First, it is recommended to draw a picture of the graph, to get an idea of how this function behaves (we notice it is a circle passing through the origin whose center is in the first quadrant. Next, let's recall the formula for arclength in polar coordinates: L = \int_{a}^{b}\sqrt{r^{2}+(\frac{dr}{d\theta })^{2}}d\theta Now, let's plug in the parameters specified in this problem: L = \int_{0}^{2\pi}\sqrt{(\sin\theta +\cos\theta)^{2}+(\cos\theta-sin\theta)^{2}}d\theta Let's FOIL everything inside the square-root sign, and we get: L = \int_{0}^{2\pi}\sqrt{\sin^{2}\theta +2\sin\theta \cos\theta +\cos^{2}\theta +\cos^{2}\theta-2\sin\theta \cos\theta+sin^{2}\theta}d\theta Now we simplify: L = \int_{0}^{2\pi}\sqrt{\sin^{2}\theta +\cos^{2}\theta +\cos^{2}\theta+sin^{2}\theta}d\theta L = \int_{0}^{2\pi}\sqrt{2(\sin^{2}\theta +\cos^{2}\theta)}d\theta Remembering our trig identity \sin^{2}\theta +\cos^{2}\theta=1, we can simplify even more: L = \int_{0}^{2\pi}\sqrt{2}d\theta Now we can integrate: L = \sqrt{2}\left [ \theta \right ]_{\theta =0}^{\theta =2\pi} L = \sqrt{2}(2\pi-0)=2\sqrt{2}\pi

### Subject: Algebra

Suzy can paint a house in 3 hours and Jim can paint the same house in 4 hours. How long would it take them to paint this house if they were working together?

We'll be using this equation to solve the problem: Speed = (amount of house painted)/(time elapsed) So, Suzy's speed = (1 house painted)/(3 hours) = \frac{1}{3} house per hour Jim's speed = (1 house painted)/(4 hours) = \frac{1}{4} house per hour Now, let's calculate their speed working together using the same concept: Speed = (amount of house painted)/(time elapsed) We'll plug in their individual speeds we just calculated: Combined Speed = (\frac{1}{3} house+ \frac{1}{4} house)/(1 hour that has passed) We can add the two fractions (we convert to a common denominator of 12), and we find: Combined Speed = \frac{7}{12} house per hour. So, now we know the speed at which they paint when working together. We're almost there! But what is the question asking for? It's asking for the amount of time it takes to the entire house. So we know the rate, and we know the amount of house being painted (the entire house...which is the same as 1 house). It turns out our same equation that we've used from the beginning has those three variables (speed, time, amount of house): Speed = (amount of house painted)/(time elapsed) Let's plug in what we know, and we get: \frac{7}{12} house per hour = (1 house painted)/(time elapsed). We need to solve for "time elapsed," so let's use our algebra (multiply both sides by time), and we get: (time elapsed)\times\frac{7}{12} house per hour = (1 house painted) Now, let's isolate "time elapsed" by dividing both sides by speed and we get: time elapsed = 1\div \frac{7}{12} house per hour time elapsed = \frac{1}{1}\times\frac{12}{7}=\frac{12}{7} = 1\frac{5}{7} hours to paint one house!

## Contact tutor

needs and Peter will reply soon.