The length of the tangent to the curve `x= a sin^(3)t, y= a cos^(3) t (a gt 0)` at an arbitrary is

Options

A. `a cos^(2)t`
B. `a sin^(2)t`
C. `(a sin^(2)t)/(cot t)`
D. `(a cos^(2) t)/(sin t)`

Answer

To find the length of the tangent to the curve defined by the parametric equations \( x = a \sin^3 t \) and \( y = a \cos^3 t \) at an arbitrary point \( t \), we can follow these steps:

### Step 1: Differentiate \( x \) and \( y \) with respect to \( t \)
We need to find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

\[ x = a \sin^3 t \implies \frac{dx}{dt} = 3a \sin^2 t \cos t \]

\[ y = a \cos^3 t \implies \frac{dy}{dt} = -3a \cos^2 t \sin t \]

### Step 2: Find \( \frac{dx}{dy} \)
Using the chain rule, we can find \( \frac{dx}{dy} \):

\[ \frac{dx}{dy} = \frac{\frac{dx}{dt}}{\frac{dy}{dt}} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} \]

This simplifies to:

\[ \frac{dx}{dy} = -\frac{\sin t}{\cos t} = -\tan t \]

### Step 3: Find the length of the tangent
The formula for the length of the tangent at a point on a curve is given by:

\[ L = |y_1| \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \]

Substituting \( y_1 = a \cos^3 t \) and \( \frac{dx}{dy} = -\tan t \):

\[ L = |a \cos^3 t| \sqrt{1 + \tan^2 t} \]

Using the identity \( 1 + \tan^2 t = \sec^2 t \):

\[ L = |a \cos^3 t| \sqrt{\sec^2 t} = |a \cos^3 t| \cdot \frac{1}{\cos t} = |a \cos^2 t| \]

Since \( a > 0 \), we can drop the absolute value:

\[ L = a \cos^2 t \]

### Final Result
Thus, the length of the tangent to the curve at an arbitrary point \( t \) is:

\[ L = a \cos^2 t \]

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