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Oishik R.

Former Test-Prep tutor, Major in Physics, Math and English

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Calculus

TutorMe

Question:

Integrate $$\int sec^3 x .\mathrm{d}x$$

Oishik R.

Answer:

Let $$I =\int sec^3 x .\mathrm{d}x $$ $$I = \int sec x. sec^2x.\mathrm{d}x$$ Integrating by parts, we get: $$I =secx \int sec^2x - \int[(\int sec^2x.dx)(\frac{\mathrm{d}secx }{\mathrm{d} x})].\mathrm{d}x$$ $$I = secx.tanx - \int(tanx).(secx.tanx).\mathrm{d}x +c_1$$ $$I = secx.tanx - \int(secx.tan^2x).\mathrm{d}x + c_1$$ $$I = secx.tanx -\int(sec^2x -1).secx.\mathrm{d}x +c_1$$ $$I = secx.tanx - \int sec^3x.\mathrm{d}x +\int secx.\mathrm{d}x +c_1$$ $$I = secx.tanx - I +ln|secx +tanx| +c_2$$ $$2I = secx.tanx + ln|secx + tanx| +c_2$$ $$I = \frac{1}{2}secx.tanx + \frac{1}{2}ln|secx + tanx| + c_3$$ Therefore: $$I =\int sec^3 x .\mathrm{d}x = \frac{1}{2}secx.tanx + \frac{1}{2}ln|secx + tanx| +k$$ where k is the constant of integration.

English

TutorMe

Question:

Evaluate critically the cosmological model presented by Aristotle in his treatise $$\textit{On the Heavens}$$.

Oishik R.

Answer:

Aristotle’s cosmology makes surprisingly few assumptions for a theory so ancient – an attribute that almost certainly contributed to its unflagging resilience. Indeed, the Aristotelian model survived almost 18 centuries, from its proposal in about 350 B.C, to the early 16th Century, when Copernicus’ ideas came to the forefront. Despite the natural tendency of Greek philosophers to disagree with their ex-mentors, Aristotle’s ideas about the universe where strikingly compatible with Plato’s, apart from a few key differences. Like most Greek philosophers of his time, Aristotle believed that the four elements – namely, earth, air, water, and fire – were the basic building blocks of all existing objects. However, he also envisioned a fifth element – that of the void, termed aether. This fifth element was different from the rest, in that it did not undergo any of Aristotle’s four changes. By definition, it was a divine material, immune to decay or degeneration. It was also possessed of a unique property, due to which it exhibited circular motion – part of the pivotal difference between Platonic and Aristotelian models..... .....But aether did not fit into this framework. Not only did it defy the forces of gravity and levity, it was unwilling to form composites – leading to the creation of celestial bodies composed solely of aether. These celestial orbs remained suspended in the spheres of the cosmos, giving off light that could be seen on the earth below. Thus we see that Aristotle did not subscribe to the Empedoclean idea of fire in the heavens, insofar as distant celestial objects were concerned. However, Aristotle most closely mirrored Plato when he described the “unmoved mover” that was at the centre of his hypothesis. The Unmoved Mover was responsible for giving rotation to the outermost sphere of the cosmos, whose motion dragged along the adjacent sphere, and so on until all the celestial orbs were spinning around the earth. It is interesting to note that early Greek orreries (most famously, the Antikythera mechanism) seem to echo this idea in their construction – all rotating dials with interlocking gears. The aether of Aristotelian cosmology can be compared to a similar transmission framework for rotational motion - further adding to its inscrutability..... (Parts of this essay have been redacted to discourage plagiarism.)

Physics

TutorMe

Question:

Show that $(\int_{s} f(\nabla\times \mathbf{A}).d\mathbf{a} = \int_{s} (\mathbf{A} \times (\nabla f)).d\mathbf{a} + \oint_{p} f\mathbf{A}.d\mathbf{l}$)

Oishik R.

Answer:

Proof: We use the product rule for gradients: $(\nabla \times (f \mathbf{A}) = f(\nabla \times \mathbf{A}) - (\mathbf{A} \times \nabla \mathit{f})$) Therefore: $(\int_{s} f(\nabla\times \mathbf{A}).d\mathbf{a} = \int_{s} \nabla \times (f \mathbf{A}).d\mathbf{a} + \int_{s} (\mathbf{A} \times \nabla \mathit{f}) $) But by Stoke's Theorem (that is, Fundamental Theory of Curls): $(\int_{s}( \nabla\times (f \mathbf{A})).d\mathbf{a} = \oint_{p} f \mathbf{A}.d\mathbf{l}$) Therefore, this implies: $( \int_{s} \nabla \times (f \mathbf{A}).d\mathbf{a} + \int_{s} (\mathbf{A} \times \nabla \mathit{f}).d\mathbf{a} = \oint_{p} f \mathbf{A}.d\mathbf{l} + \int_{s} (\mathbf{A} \times \nabla \mathit{f}).d\mathbf{a}$) Which is the required statement.

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