Question

A circle of radius 2 and center \((0,0)\) can be parametrized by the equations \(x=2 \cos t\) and \(y=2 \sin t .\) Show that for any value of \(t\) , the line tangent to the circle at \((2 \cos t, 2 \sin t)\) is perpendicular to the radius.

Short answer

The tangent line to the circle at the point \((2 \cos t, 2 \sin t)\) is indeed perpendicular to the radius. This claim was verified using the properties of derivatives (which provide the slope of the tangent line) and the geometric condition that two lines are perpendicular if the product of their slopes is -1, both of which held true in this case.

Step-by-step solution

Compute the derivative of x and y with respect to t

The derivative of \(x = 2 \cos t\) with respect to \(t\) is \(-2 \sin t\). Similarly, derive \(y = 2 \sin t\) to get \(2 \cos t\). These derivatives represent the slopes of the tangent line.

Compute the slope of the radius

Since the radius of the circle lies along the line connecting the origin (0,0) to the point on the circle, its slope is given by \(\frac{y - 0}{x - 0} = \frac{2 \sin t}{2 \cos t} = \tan t\).

Check for the perpendicular condition

Two lines are perpendicular if the product of their slopes is -1. The slope of the tangent line is given by \(\frac{dy}{dx} = \frac{2 \cos t}{-2 \sin t} = - \cot t\). Therefore, the product of the slopes is \(- \cot t \times \tan t = -1\). As the product of the slopes is -1, the tangent line and the radius are orthogonal.