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 3 months ago
can someone help me on the question 2C10. It
is Unit 2 Exercises Problem Set 4. The question can be found at this link http://ocw.mit.edu/courses/mathematics/1801scsinglevariablecalculusfall2010/unit2applicationsofdifferentiation/partboptimizationrelatedratesandnewtonsmethod/problemset4/MIT18_01SC_pset2prb.pdf. I have problems coming up with the basic function to represent time.
 3 months ago
can someone help me on the question 2C10. It is Unit 2 Exercises Problem Set 4. The question can be found at this link http://ocw.mit.edu/courses/mathematics/1801scsinglevariablecalculusfall2010/unit2applicationsofdifferentiation/partboptimizationrelatedratesandnewtonsmethod/problemset4/MIT18_01SC_pset2prb.pdf. I have problems coming up with the basic function to represent time.

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phi
 3 months ago
Best ResponseYou've already chosen the best response.0The solution is posted... However, they did manage to confuse the sin with the cosine The relation you will find for this problem is \[ \frac{\cos \alpha}{\cos{\beta}} = \frac{5}{2} \] In Snell's Law, they define the angle with respect to the surface normal, i.e. 90\(\alpha\)), and you end up with sin (rather than cos) as for setting up the equations, the most basic idea is rate * time = distance and time= distance/rate Use Pythagoras to find the length of the hypotenuse of the triangle (from A to the water line). Divide this distance by the speed = 5 m/s Call this t1 Do the same for the other triangle, but use speed = 2 m/s, and call this t2 Total time is t1+t2, and it is a function of x (distance from pt P, where you dive into the water): \[ T(x) = t_1(x) + t_2(x) \] Take the derivative with respect to x, and set equal to zero, to find the critical point: \[\frac{d}{dx} T(x) = 0 \\\frac{d}{dx}t_1(x) + \frac{d}{dx}t_2(x) = 0 \]
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