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Tutor profile: Ben H.

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Ben H.
Math and Science nerd; teacher at heart
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Questions

Subject: Microsoft Excel

TutorMe
Question:

What is the difference between the COUNT, COUNTA, COUNTIF and COUNTBLANK formulas?

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Ben H.
Answer:

COUNT is used to count cells containing any numerical value and will exclude blank cells. COUNTA or Count All is used to count any cell value containing more than numerical values including text and boolean. This formula will also ignore blank cells. COUNTBLANK will return the sum of any blank cell or cells containing a blank string COUNTIF and COUNTIFS count cells based on certain criteria, COUNTIF contains one dependency while COUNTIFS allows the user to use or nest multiple criteria.

Subject: Calculus

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Question:

What is the area under the surface $$ z = 5 - x^{2} + 4y $$ over the region $$ \mathcal{D} $$ defined by $$ 0 \leq x \leq 1 $$ and $$ -x \leq y \leq x $$?

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Ben H.
Answer:

Use a double integral to solve for the area under the equation over the region $$ \mathcal{D} $$. $$ \int\int_\mathcal{D} (5 - x^{2} + 4y) \mathrm{d}x \mathrm{d}y $$ The region defines the boundaries so we have $$ \int_0^1\int_{-x}^x (5 - x^{2} + 4y) \mathrm{d}x \mathrm{d}y $$ Integrating the first, $$ \int_0^1 (5y - x^{2}y + \frac{4y^{2}}{2}) \mathrm{d}x \Big\rvert_{-x}^{x} $$ $$ = \int_0^1 (10x - 2x^{3}) \mathrm{d}x $$ Which will integrate to $$ (5x^{2} - \frac{1}{2}x^{4}) \Big\rvert_{0}^{1} $$ Evaluating according to the boundary conditions gives $$ Area = 4.5 $$

Subject: Physics

TutorMe
Question:

What is the Lagrangian for a particle near the surface of the earth confined to the parabolic curve $$ y = ax^{2} $$, where $$ y $$ is up?

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Ben H.
Answer:

The canonical Lagrangian can be expressed as: $$ \mathcal{L}(q,\dot{q},t) = T - U, $$ where $$ T $$ is the kinetic energy and $$ U $$ is the potential energy. For a particle near the surface of the earth, $$ T = \frac{1}{2}mv^{2} $$, and $$ U = mgh $$, where $$ v^{2} = (\dot{x}^{2} + \dot{y}^{2}) $$ and $$ h =y$$ Giving, $$ T - U= \frac{1}{2}m(\dot{x}^{2} + \dot{y}^{2}) - mgy $$. Thus we must determine the corresponding $$ \dot{q} $$ coordinates. $$ \dot{y} = \frac{d}{dt}ax^{2} = 2a\dot{x} $$ Using this result, we get $$ \dot{x} = \frac{\dot{y}}{2a} $$, and $$ \dot{x}^{2} = \frac{\dot{y}^{2}}{4a^{2}x^{2}} = \frac{\dot{y}^{2}}{4ay}$$ Returning to our Lagrangian and replacing coordinates, we have $$ \mathcal{L} = \frac{1}{2}m(\frac{\dot{y}^{2}}{4ay} + \dot{y}^{2}) - mhy $$ Which simplifies to $$ \mathcal{L} = \frac{1}{2}m\dot{y}^{2}(1 +\frac{1}{4ay}) - mhy $$

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