Express a Function in Terms of Another FunctionDate: 09/18/97 at 13:29:20 From: Migdalia Anes Subject: Expressing a function in terms of another function The altitude perpendicular to the hypotenuse of a right triangle is 12 cm. Express the length of the hypotenuse in terms of the perimeter. I tried lots of things, but they're wrong. I really don't know how to get it! Date: 09/24/97 at 17:08:08 From: Doctor Rob Subject: Re: Expressing a function in terms of another function Draw a diagram of a right triangle with hypotenuse h, perimeter p, and one leg x, other leg p-h-x: . /| / | / | / | / | h / | p-h-x / | / | /`-.12 | / `-. | /________`-| x Applying the Pythagorean Theorem to the small triangles gives the two parts of the hypotenuse as Sqrt[x^2-144] and Sqrt[(p-h-x)^2-144]. Their sum must be h. The Pythagorean Theorem on the large triangle gives another equation. Together we have: x^2 + (p-h-x)^2 = h^2, h = Sqrt[x^2-144] + Sqrt[(p-h-x)^2-144]. To remove the radicals from the second equation, rewrite it as h - Sqrt[x^2-144] = Sqrt[(p-h-x)^2-144], and square both sides: h^2 - 2*h*Sqrt[x^2-144] + x^2 - 144 = (p-h-x)^2 - 144. Add 144 to both sides, and replace (p-h-x)^2 by h^2 - x^2 from the first equation. h^2 - 2*h*Sqrt[x^2-144] + x^2 = h^2 - x^2. Isolating the radical on the right side, you get x^2 = h*Sqrt[x^2-144]. Now square both sides again, and the radicals disappear: x^4 = h^2*(x^2-144). Combine this with the first equation, and you get the system x^4 - h^2*x^2 + 144*h^2 = 0, x^2 + (p-h-x)^2 - h^2 = 0. You want to eliminate x from these equations, to get one equation involving h and p alone. Expand the second equation: x^2 + p^2 + h^2 + x^2 - 2*h*p - 2*p*x + 2*h*x - h^2 = 0, 2*x^2 + (2*h-2*p)*x + (p^2-2*h*p) = 0. Find the quotient Q of the quartic polynomial divided by the quadratic one as polynomials in x. Subtract Q times the quadratic one from the quartic one. That will leave, fortuitously, a quadratic equation in h. Factor it, and set each factor linear in h equal to zero. These can be solved for h to give the desired expressions. Of course you have to check your answer, since squaring twice could have introduced spurious solutions! Actually, you can use the known fact that h < p/2 to throw out one of the solutions. Once you find h in terms of p, there will be two values of x for each value of p, because we didn't specify if x was the longer leg or the shorter one. Thus x will be the root of a quadratic equation, namely the last one given above, with coefficients functions of p. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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