# Tutor profile: Ryan S.

## Questions

### Subject: Calculus

A company's sales over the course of a year have been found to be quite well modeled by the function $$S(t) = -4t^2+52t+120$$, where $$S$$ represents the sales at a given point in time in units of thousands of products, and $$t$$ represents time of year in units of months. During what month of the year does this company experience its maximum in sales?

The maximum of a function can be found by determining where the function's derivative is equal to 0 and its second derivative is negative. Looking at the function provided for the company's sales, its derivative is $$ S'(t) = -8t + 52$$ . So, setting this equal to 0: $(-8t+52=0 \implies 8t=52 \implies t=6.5$) .Moreover, checking its second derivative at that time: $(S''(t) =-8\\ S''(6.5) = -8$). Thus, because the first derivative is equal to 0 and the second derivative is negative then, it can be concluded that the company reports its maximum in sales at $$t=6.5$$, or in June.

### Subject: Astronomy

Earth's orbital eccentricity is ~0.02 and its axis tilt is ~23.5 degrees relative to the plane of its orbit around the Sun. Why does Earth’s orbital eccentricity not matter when considering seasons?

Earth’s orbital eccentricity is very small at 0.02 ($$0.02<<1$$), meaning that Earth’s orbit around the Sun is quite close to being perfectly circular, and as a result, Earth experiences barely any noticeable changes due to variations in distance over the course of its orbit. On the other hand, Earth’s axis tilt is fairly large, and the variation in the directness of sunlight for the 2 hemispheres is what really causes seasons.

### Subject: Physics

A satellite orbits Earth with a speed of 11,000 kph. How much more slowly (as a percentage) does a clock onboard this satellite tick as observed from Earth's surface compared to this same clock observed from inside the satellite?

According to the theory of special relativity, the clock will appear to tick more slowly when observed by someone on Earth's surface than it will when observed by someone onboard the satellite--this is the effect of time dilation. So, the equation used is the time dilation equation, $$\Delta t = \gamma \Delta t_0$$, where $$\Delta t$$ is the time elapsed observed from Earth, $$\Delta t_0$$ is the time elapsed observed from onboard the satellite, and $$\gamma$$ is the Lorentz factor: $(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} $), with $$v$$ the relative speed and $$c$$ the speed of light. The question asks much MORE slowly the clock is observed to tick. So, this change as a decimal would be $$\frac{\Delta t - \Delta t_0}{\Delta t_0}$$, or $$\gamma - 1$$. The speed of light in kph is approximately $$1.1 \times 10^9$$. So, plugging in the numbers, we get that $$\gamma \approx 1.00000000005$$. Thus, as a decimal, the clock ticks more slowly with a change of $$\gamma - 1 = 0.00000000005$$, or, as a percentage, 0.000000005% more slowly--barely any more slowly, even with a satellite orbiting Earth quite fast.

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