Question

The length of the tangent to the curve $x = a \left( \cos t + \log \tan \frac{t}{2} \right), y = a \sin t$ is (where $0 < t < \frac{\pi}{4}$)

• A. ax
• B. ay
• C. a
• D. xy

Answer 1

Answer: (C)

a

Answer 2

The parametric equations of the curve are given by $x = a \left( \cos t + \log \tan \frac{t}{2} \right)$ and $y = a \sin t$, where $0 < t < \frac{\pi}{4}$.

First, we find the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$. $\frac{dx}{dt} = \frac{d}{dt} \left[ a \left( \cos t + \log \tan \frac{t}{2} \right) \right]$ $\frac{dx}{dt} = a \left( -\sin t + \frac{1}{\tan \frac{t}{2}} \cdot \sec^2 \frac{t}{2} \cdot \frac{1}{2} \right)$ Using $\tan \frac{t}{2} = \frac{\sin \frac{t}{2}}{\cos \frac{t}{2}}$ and $\sec^2 \frac{t}{2} = \frac{1}{\cos^2 \frac{t}{2}}$, we get: $\frac{dx}{dt} = a \left( -\sin t + \frac{\cos \frac{t}{2}}{\sin \frac{t}{2}} \cdot \frac{1}{\cos^2 \frac{t}{2}} \cdot \frac{1}{2} \right)$ $\frac{dx}{dt} = a \left( -\sin t + \frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}} \right)$ Using the double angle formula $\sin t = 2 \sin \frac{t}{2} \cos \frac{t}{2}$, we have: $\frac{dx}{dt} = a \left( -\sin t + \frac{1}{\sin t} \right) = a \left( \frac{1 - \sin^2 t}{\sin t} \right) = a \frac{\cos^2 t}{\sin t}$.

$\frac{dy}{dt} = \frac{d}{dt} (a \sin t) = a \cos t$.

The slope of the tangent to the curve at parameter $t$ is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. $\frac{dy}{dx} = \frac{a \cos t}{a \frac{\cos^2 t}{\sin t}} = \frac{\cos t \sin t}{\cos^2 t} = \frac{\sin t}{\cos t} = \tan t$.

Let the point of tangency be $(x, y)$. The equation of the tangent line at this point is $Y - y = \frac{dy}{dx} (X - x)$, which is $Y - y = \tan t (X - x)$.

The "length of the tangent" usually refers to the length of the segment of the tangent line from the point of tangency to its intersection with the x-axis or the y-axis. Let's calculate both.

1. Length of the tangent segment from $(x, y)$ to the x-axis: The tangent intersects the x-axis when $Y=0$. $0 - y = \tan t (X - x)$ $-y = \tan t (X - x)$ $X - x = -\frac{y}{\tan t} = -y \cot t$ $X = x - y \cot t$. The intersection point is $(x - y \cot t, 0)$. The length of the tangent segment $L_x$ is the distance between $(x, y)$ and $(x - y \cot t, 0)$. $L_x = \sqrt{((x - y \cot t) - x)^2 + (0 - y)^2} = \sqrt{(-y \cot t)^2 + (-y)^2}$ $L_x = \sqrt{y^2 \cot^2 t + y^2} = \sqrt{y^2 (\cot^2 t + 1)} = \sqrt{y^2 \csc^2 t} = |y \csc t|$. Substitute $y = a \sin t$: $L_x = |a \sin t \cdot \csc t| = |a \sin t \cdot \frac{1}{\sin t}| = |a|$. Since $a$ is a scale factor representing length, it is usually taken as positive. So, $L_x = a$.

2. Length of the tangent segment from $(x, y)$ to the y-axis: The tangent intersects the y-axis when $X=0$. $Y - y = \tan t (0 - x)$ $Y - y = -x \tan t$ $Y = y - x \tan t$. The intersection point is $(0, y - x \tan t)$. The length of the tangent segment $L_y$ is the distance between $(x, y)$ and $(0, y - x \tan t)$. $L_y = \sqrt{(0 - x)^2 + ((y - x \tan t) - y)^2} = \sqrt{(-x)^2 + (-x \tan t)^2}$ $L_y = \sqrt{x^2 + x^2 \tan^2 t} = \sqrt{x^2 (1 + \tan^2 t)} = \sqrt{x^2 \sec^2 t} = |x \sec t|$. Substitute $x = a \left( \cos t + \log \tan \frac{t}{2} \right)$: $L_y = \left| a \left( \cos t + \log \tan \frac{t}{2} \right) \sec t \right| = \left| a \left( \frac{\cos t}{\cos t} + \frac{\log \tan \frac{t}{2}}{\cos t} \right) \right| = \left| a \left( 1 + \frac{\log \tan \frac{t}{2}}{\cos t} \right) \right|$. This expression does not simplify to one of the options $ax, ay, a, xy$.

Since one of the standard interpretations of "length of the tangent" (the length of the segment from the point of tangency to the x-axis) yields the result 'a', which is one of the options, this is likely the intended meaning of the question.