A boy standing at corner A of a rectangular pool has a boat in corner B on the e

Question

A boy standing at corner A of a rectangular pool has a boat in corner B on the end of a string that is 30 feet long. He walks along the side of the pool toward C, keeping the string taunt. a) Find an equation of the boat?s path. That is, find a relationship between x and y. Hints: Consider the right triangle with hypotenuse PE. Note that the tangent of ? can be expressed in terms of the side lengths of this triangle and that dy dx = tan? . You might also find that it is easier to solve this DE for x than for y. You may use a CAS to do the integration. b) What is the distance from AB to the boat when the boat is half way across the pool? How far has the boy walked at that time?

Explanation / Answer

It is hinted that dy/dx = tan(theta). We use the triangle given in the diagram and since y = 30, using the pythagorean theorem, for tan(theta) we get the DE or dy/dx = - y / sqrt( 900 - y^2). Where, sine = y and cosine = sqrt( 900 - y^2 ), remember tan = sin/cos. It is also hinted that you want to solve for x rather than y. This will help for part b. Knowing that dy/dx = 1/(dy/dx). We rewrite our DE in the form... dx/dy = - sqrt( 900 - y^2 ) / y. Now we seperate and integrate... to get... x(y) = 30*ln( sqrt( 900-y^2-30 )/y ) + sqrt(900 - y^2) + C Given that x(30) = 0, we use this as a initial condition...the string 30 feet long at A, or y = 30 at x = 0. We set the equation written above of x(y) = 30 and solve for C. We get C = - 30*ln ( sqrt(900 - y^2 - 30)/30 ), we then simplify this to get... x(y) = 30*(ln/y) + sqrt(900-y^2) Taking half way across the pool to mean an initial condition of x(15)= ? , 30/2 = 15 we plug in 15 for y into the simplified x(y) to get ... The distance of the boat from AB at half away across sthe pool to be 46.7752 ft or 46.78 ft.

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