Earl

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These are questions asked by Earl

PrimesQuestion.mw

Please let me know if this link correctly accesses my worksheet. If not, I will copy its contents into this question.

Which ODE in the worksheet, if any, provides the correct answer?


restart

f := proc (x) local t; if not type(evalf(x), 'numeric') then ('procname')(x) else evalf(Int(exp(-(1/10)*t^2), t = 0 .. x)) end if end proc

solA := dsolve({diff(y(x), x) = y(x)+f(x), y(0) = 0}, numeric, known = f)

solA(1)

[x = 1., y(x) = HFloat(0.7081492947996167)]

(1)

f2 := evalf(Int(exp(-(1/10)*t^2), t = 0 .. 1)); f(1)

.9676433126

 

.9676433126

(2)

solB := dsolve({diff(y(x), x) = y(x)+f2, y(0) = 0}, numeric, output = listprocedure)

solB(1)

[x(1) = 1., (y(x))(1) = HFloat(1.6626837619970016)]

(3)

YinSolB := subs(solB, y(x))

YinSolBeval := solve(YinSolB(a) = .7081, a); solB(YinSolBeval)

.5491485953

 

[x(.5491485953) = .5491485953, (y(x))(.5491485953) = HFloat(0.7081000000284681)]

(4)

NULL


 

Is there a help page which explains why braces provide the partial text evaluation in this code?

RopeLen := 30;RopeAddLen := .5;

plots[textplot]([1, 1, typeset("%1", (({(1/2)*RopeLen}+{(1/2)*RopeAddLen})^2-{(1/2)*RopeLen}^2)^(1/2))])

eq1 := z = y*log(x): eq2 := z = y+x*log(x):

DispIntersecting := implicitplot3d([eq1, eq2], x = 0 .. 10, y = -30 .. 30, z = -40 .. 40, color = [blue, green]):

solve({eq1, eq2}, [x, y, z]);

assign(%):

DispIntersection := spacecurve([x, y, z], x = 0.1e-2 .. 10, color = red, view = [0 .. 10, -30 .. 30, -40 .. 40]):

display(DispIntersecting, DispIntersection, axes = boxed, scaling = constrained);

From a Maple Primes answer two years ago:

f(x,y) is the equation of a line through point [m,n]. The solve command finds values of a and b for which f(x,y) are lines through [m,n] and tangent to x^2 + y^2 = r^2.

f := proc (x, y) options operator, arrow; a*(x-m)+b*(y-n) end proc

solve([f(0, 0) = r, a^2+b^2 = 1], [a, b])

These commands are far from the conventional solution. Why do they provide the correct answers?

Has anyone solved this problem from an older Putnam paper?

An ellipse sitting in the first quadrant with its major axis parallel to the x axis is tangent to the positive x and y axes.

It slides clockwise within the first quadrant while maintaining tangency to both positive axes until its major axis is parallel to the y axis.

Prove that the locus of its centre is the arc of a circle.

I have crudely animated this motion by sliding the axes around the stationary ellipse. Is there a more elegant animation which slides the ellipse against stationary axes?

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