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Chapter 8: Problem 22

Find the equations of the tangent lines at the point where the curve crosses itself. $$ x=2-\pi \cos t, \quad y=2 t-\pi \sin t $$

Short Answer

Expert verified
The equations of the tangent lines to the points where the curve crosses itself are given by the equation(s) of the line(s) obtained in Step 4. The exact answers will depend upon the value(s) of 't' that satisfy the equation from Step 1.

Step by step solution

01

Solving for 't'

Set the equations for 'x' and 'y' equal to each other, i.e., \(2 - \pi \cos t = 2t - \pi \sin t\), and solve for 't'. Since this can be a complex task due to the trigonometric terms, we can graphically or numerically solve for the possible value(s) of 't'.
02

Find the Point(s) of Intersection

Substitute the value(s) of 't' from Step 1 into the parametric equations to find the corresponding 'x' and 'y' coordinates, i.e., \(x = 2 - \pi \cos t_\text{sol}\) and \(y = 2t_\text{sol} - \pi \sin t_\text{sol}\). This will give us the point(s) where the curve crosses itself.
03

Determining the gradients of the Tangent lines

Differentiate the parametric equations with respect to 't' to get \(\frac{dx}{dt} = \pi \sin t\) and \(\frac{dy}{dt} = 2 - \pi \cos t\). Then find \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) which will give the gradient of the tangent line at any point on the curve. Substitute the 't' value(s) found in Step 1 into \(\frac{dy}{dx}\) to get the gradient(s) of the tangent line(s) at the intersection point(s).
04

Writing the Equations of the Tangent Lines

Using the point-slope form of a linear equation which is \(y - y_1 = m(x - x_1)\), where \(m\) is the gradient of the line and \((x_1, y_1)\) is a point on the line, we can write the equation of the tangent line(s) at the intersection point(s). Substitute the coordinates of the intersection point(s) found in Step 2 and their respective gradient(s) from Step 3 into the point-slope form to get the equation(s) of the tangent line(s).

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