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# Tutor profile: William B.

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William B.
Tutor for 6 years
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## Questions

### Subject:Pre-Calculus

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Question:

What is the sum of the following series when n goes towards infinity, $$S = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ... + \frac{1}{n}$$

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William B.

A simple solution to this problem is to take 2S and substract S. $$2S = 1+ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ... + \frac{2}{n}$$ $$S = 2S - S = 1 + 1/n$$ Which simplifies when n goes towards infinity to, $$S=1+0=1$$

### Subject:Basic Math

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Question:

If Jana has 5 apples and John has 20 apples, how many apples does John have to give to Jana for them the same exact amount of entire apples. If it is impossible, why and what's the minimum amount of apples they have to eat for them to be able to share the same amount of apples?

Inactive
William B.

First, we find the total amount of apples. $$5+20 = 25$$ Next, we find what half of this total amount is. $$25/2 = 12.5$$ Since we obtain a decimal number, it is impossible for Jana and John to have the same exact amount of entire apples! They would need to split an apple in half to have the same amount. Hence, the maximum apples they can both have is 12. This amounts to a total of 24 apples they can equally share, leaving 1 apple that they have to eat.

### Subject:Calculus

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Question:

The vector field $$F(x,y) = (2xe^y + 1, x^2e^y )$$ is conservative. Compute the work done by the field in moving an object along the path $$C: r(t) = (cos(t),sin(t)), 0 \leq t \leq pi$$.

Inactive
William B.

Start out by substituting x and y into the vector field equation F by cos(t) and sin(t) respectively. $$F(t) = (2cos(t)*e^{sin(t)} + 1, cos^2(t)*e^{sin(t)})$$ The equation to find the work is the following. $$W = \int_{c}F dr\$$ We are missing $$dr$$ which can be found by derivating r(t). $$dr = (-sin(t),cos(t))dt$$ Substituting everything in, we obtain the following equation. $$W = \int_{0}^{\pi}\ (2cos(t)*e^{sin(t)} + 1, cos^2(t)*e^{sin(t)})*(-sin(t),cos(t))dt$$ $$W = \int_{0}^{\pi}\ ((2cos(t)*e^{sin(t)} + 1)(-sin(t))+(cos^2(t)*e^{sin(t)})*cos(t))dt$$ $$W = \int_{0}^{\pi}\ (-2sin(t)*cos(t)*e^{sin(t)} - sin(t) +cos^2(t)*e^{sin(t)}*cos(t))dt$$ That integral can be quite scary, so we'll separate it into smaller integrals and solve it piecewise. $$W = \int_{0}^{\pi}\ (-2sin(t)*cos(t)*e^{sin(t)})dt + \int_{0}^{\pi}\ -sin(t)dt + \int_{0}^{\pi}\ cos^2(t)*e^{sin(t)}*cos(t)dt$$ $$W = A + B + C$$ where, $$A = \int_{0}^{\pi}\ (-2sin(t)*cos(t)*e^{sin(t)})dt$$ $$B = \int_{0}^{\pi}\ -sin(t)dt$$ $$C = \int_{0}^{\pi}\ cos^2(t)*e^{sin(t)}*cos(t)dt$$ $$\textbf{A}$$ $$A = \int_{0}^{\pi}\ (-2sin(t)*cos(t)*e^{sin(t)})dt$$ For A, we'll use the substitution, $$u = sin (t)$$ $$\frac{du}{dt}= cost(t)$$ $$A=-2\int_c ue^udu$$ $$\textbf{B}$$ $$B = \int_{0}^{\pi}\ -sin(t)dt$$ Which is simply, $$B =cos(t)$$ $$\textbf{C}$$ $$C = \int_{0}^{\pi}\ cos^2(t)*e^{sin(t)}*cos(t)dt$$ Using the same substitution as in A, $$u = sin (t)$$ $$\frac{du}{dt}= cost(t)$$ We obtain, $$C= \int_c (1-u^2)e^udu$$ $$C= \int_c e^udu - \int_c u^2e^udu$$ $$C= e^u - \int_c u^2e^udu$$ Using, $$v=u^2$$ $$\frac{dv}{du} = 2u$$ and, $$dw = e^u$$ $$w = e^u$$ We can use the theorem, $$\int_c vwdu = vw - \int_c wdv$$ Giving, $$C= e^u - (u^2e^u - 2\int_c ue^udu)$$ $$C= e^u - u^2e^u + 2\int_c ue^udu$$ Adding everything together, we now obtain, $$W = A + B + C$$ $$W = [-2\int_c ue^udu + cos(t) + e^u - u^2e^u + 2\int_c ue^udu] \Big|_0^\pi$$ A cancels out with the third component of C, simplifying W to, $$W = [cos(t) + e^u - u^2e^u ] \Big|_0^\pi$$ $$W(t) = [cos(t) + e^{sin(t)} - sin^2(t)e^{sin(t)} ] \Big|_0^\pi$$ $$W = [(0+e-e)-(1+1-0)]$$ $$W = -2$$ The work done by the field is found to be equal to -2.

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