Problem 14 - Algebra and Trigonometry

Question

Write the appropriate rotation formulas so that in a rotated system the equation has no \(x^{\prime} y^{\prime}\) -term. $$32 x^{2}-48 x y+18 y^{2}-15 x-20 y=0$$

Step-by-Step Solution

Answer

The new variables \(x'\) and \(y'\) will be \(x' = x \cos(\theta) - y \sin(\theta)\) and \(y'=x \sin(\theta) + y \cos(\theta)\) respectively. When you substitute and simplify, the rotated equation will be obtained with no \(x'y'\) term.

Step 1: Identify Coefficients

Identify the coefficients of \(x^2\), \(y^2\) and \(xy\) terms i.e \(A = 32\), \(B = -48\) and \(C = 18\) respectively from the given equation.

Step 2: Determining Angle of Rotation

The angle of rotation \(\theta\) can be calculated by using the formula \(\tan(2\theta) = \frac{B}{A-C}\). Substituting the values, we get \(\tan(2\theta) = \frac{-48}{32 - 18} = -3\). Solving for \(\theta\) gives \(\theta = \frac{1}{2} \tan^{-1}(-3)\).

Step 3: Form New Equation

The rotated equation will now be in the form of new variables \(x'\) and \(y'\). The formulas for \(x'\) and \(y'\) in terms of \(x\) and \(y\) are given by: \(x' = x \cos(\theta) - y \sin(\theta)\) and \(y' = x \sin(\theta) + y \cos(\theta)\). Substituting these into the original equation will give a new rotated equation without the \(x'y'\) term.