Question

Question 2.5 and 2.6
Good day professor

I've manage to do most of it but just got stuck with 2.5 and 2.6! So the assistance would be appreciate it! Many thanks :)

I've attached a link to the question and a picture!

For Probs 2.5 and 2.6, we can assume you have already solved for the constants a, b, c for the parabola f(x) and m and y0 for the line g(x) given by

g(x) = mx + y0.

We can also assume that you have figured out the coordinate of the midpoint of g(x), i.e., (xm, ym)

For 2.5, note that a hyperbola has y ~ 1/x and in general can be written

h(x) = A/(x-x∞)

where A is an overall scaling factor and x∞ is the value of x where the hyperbola goes off to infinity. Since we have 2 unknowns, A and x∞, but only one eqn between them, namely h(xm) = ym, coupled with the fact that xm < 0 so that x∞ cannot be 0, it looks lie we need to assume a value for A. So let A = 1.

So now just set

h(xm) = ym = 1/(xm - x∞)

and solve for x∞.

For 2.6, we need to find the maximum of a function z(x) = distance between f and g. Finding an extremum, such as a maximum, involves setting the derivative equal to zero, so first find the eqn for z(x) and take the derivative. So

z(x) = ax^2 + bx + c - mx - y0

and

dz/dx = 2ax + b - m = 0

and solve for x. (  I get x = (m - b)/2a ).

Good luck!
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 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Thank you for your time, appreciate the assistance :)

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#### randy patton

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college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography

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26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related

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J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

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M.S. MIT Physical Oceanography, B.S. UC Berkeley Applied Math

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