• Knowledge
can someone help me on the question 2C-10. It is Unit 2 Exercises Problem Set 4. The question can be found at this link I have problems coming up with the basic function to represent time.
MIT 18.01 Single Variable Calculus (OCW)
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • jamiebookeater
I got my questions answered at in under 10 minutes. Go to now for free help!
  • phi
The 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: \[\frac{d}{dx} T(x) = 0 \\\frac{d}{dx}t_1(x) + \frac{d}{dx}t_2(x) = 0 \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.