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# Tutor profile: Alexander C.

Alexander C.
Tutor for 3 years and Graduate Student in Physics

## Questions

### Subject:Calculus

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Question:

Evaluate the integral, $$A = \int\frac{1}{x^2+x}dx$$.

Alexander C.

The bottom of the fraction simplifies to $$x(x+1)$$. We can now separate the denominator then into two partial fractions and then integrate, $$A = \int(\frac{1}{x} - \frac{1}{x+1})dx = \ln(x) - \ln(x+1) + c = \ln(\frac{x}{x+1}) + c$$, where we used the $$\ln(x)$$ rule namely, $$\ln(a) - \ln(b) = \ln(\frac{a}{b})$$.

### Subject:Physics

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Question:

Suppose that the distance between two plates of a capacitor is $$2.0mm$$ with an area of $$4.5\times 10^{-3}m^2$$. Then, determine the electrical field between the plates before and after some $$Teflon^{TM}$$ is inserted if $$V_0=40 V$$.

Alexander C.

The electrical field is given by $$E_0$$ as $$E_0 = \frac{V_0}{d} = \frac{40V}{2.0\times 10^{-3}m} = 2.0\times 10^4 V/m$$. The true electric field then with the $$Teflon^{TM}$$ in place is $$E = \frac{1}{\kappa}E_0 = \frac{1}{2.1}2.0\times 10^4 V/m = 9.5\times 10^3 V/m$$, where the $2.1$ comes from a table that can be commonly found for dielectric constants.

### Subject:Differential Equations

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Question:

Given the following differential equation, find the general solution (no initial conditions given): $$\frac{dy}{dx} = x+xy$$.

Alexander C.

We solve by using seperation of variables. First, factor out the $$x$$ to get, $$(1+y)x$$ on the right-hand side, then divide out the $$dx$$ differential to obtain, $$\int\frac{dy}{(1+y)} = \int xdx$$. Integrating both sides using our $$\ln(x)$$ rules we find, $$\ln(1+y) = \frac{x^2}{2} + c$$ where $$c$$ is some arbitrary constant. Solving for $$y$$ we find, $$y(x) = A\exp(\frac{x^2}{2})$$, where we absorbed all the constants into the new $$A$$ value.

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