# Tutor profile: Vinay R.

## Questions

### Subject: Pre-Calculus

Convert the following equation into polar coordinates: (x^2)/a^2 + (y^2)/b^2 = 1

You may recognize this equation as the equation of an ellipse and perhaps know the equivalent polar equation. We will derive it anyway. By definition: x = rcos(θ), y=rsin(θ). Substituting into the original equation: r^2cos^2(θ)/a^2 + r^2sin^2(θ) = 1 Factoring out r^2: r^2(cos^2(θ)/a^2 + sin^2(θ)/b^2) = 1 Then: r^2 = 1/(cos^2(θ)/a^2 + sin^2(θ)/b^2) Simplifying: r^2 = a^2*b^2/(b^2cos^2(θ)+a^2sin^2(θ)) <-- final answer This looks messy but let's say a and b are equal to c (in the case of a circle as you may know): r^2 = c^4/(c^2cos^2(θ)+c^2sin^2(θ) Factoring out and cancelling the c^2: r^2 = c^2/(cos^2(θ)+sin^2(θ)). From trig identities cos^2(θ)+sin^2(θ) = 1! So, r=c (sign doesn't really matter in this case because we are not concerned with direction of any sort). This is the equation of a circle in polar!

### Subject: Calculus

An object is moving along the x-axis with velocity defined by v(t)=tsin(t) over the time interval 0 to 2π. At t=π, x=0. Find the total distance traveled over this time interval. Find the position at t=0.

The distance over a time interval is the integral of the absolute value of the velocity over the time interval. This is because distance does not depend on direction. Since v(t) changes sign over the interval 0 to 2π, we must split this integral into two parts: when v(t) > 0 (from 0 to π) and when v(t) < 0 (from π to 2π). Both integrals will involve finding the antiderivative of tsin(t) which we must find using integration by parts. The general formula of integration by parts is integral of udv is uv - integral of vdu. Let u=t and dv=sin(t). Then the integral of tsin(t)dt (which we want to find) is equal to -tcos(t) - integral -cos(t)dt which equals: sin(t) - tcos(t) + C. Integral 1 from 0 to π when evaluating comes to [0+π] - [0-0] = π. Integral 2 from 0 to 2π when evaluating comes to | [0-2π] - [0+π] | = 3π. Distance is the sum of the two integrals: 3π. To find the initial value, we are given that at t=π, x=0. Plug this into the antiderivative we get: 0 = 0 + π + C. Then C=-π. Plugging in t=0, we find that x= -π. Another way to solve this second step is that you found integral 1 from 0 to π. You know that velocity is positive here so x is increasing. The result (π) has to be equal to x(π)-x(0). x(π) = 0 so 0-x(0)=π. Therefore x(0)=-π.

### Subject: Algebra

A man gives you a bag of coins. He says, "If you tell me the number of dimes that are in this bag I will give it to you." He gives you the following pieces of information: 1. In the bag there are only quarters, dimes and nickels and there are 100 coins total. 2. The value of all the coins in total is $11.20 3. The value of the dimes in the bag exceeds the value of nickels in the bag by 95¢

Let q be the number of quarters in the bag. Let d be the number of dimes in the bag. Let n be the number of nickels in the bag. 1. q + d + n = 100 (there are 100 coins in the bag and there are only quarters dimes and nickels) 2. 0.25q + 0.10d + 0.05n = 11.20 (the total value of the coins sum to $11.20. The total value for each coin is calculated by the value of an individual coin - 0.25 for quarters, 0.10 for dimes, and 0.05 for nickel - multiplied by the number of that particular coin) 3. 0.10d - 0.05n = 0.95 (the total value of dimes minus the total value of nickels is $0.95) 4. q + 3d = 119 (Add equation 1 to 20 times equation 3 to eliminate n) 5. 0.25q + 0.2d = 12.15 (add equation 2 to equation 3 to eliminate n) 6. 2.2d = 70.40 (Subtract 4 times equation 5 from equation 4 to eliminate q) 7. d = 32 (Divide both sides by 2.2) The question simply asks for us to find the number of dimes in the bag (32 we just found). If we wanted to find the number of quarters or nickels we just have to plug in this value into equation 3 to find nickels and then plug in both those values into equation 1 or 2. If you wish to work out the number of nickels and quarters you will find that there are 23 quarters, 32 dimes, and 45 nickels.

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