The length of the normal at `theta` on the curve `x = a cos^3 theta,y=asin^3theta` is

Options

A. `asin^2theta`
B. `asin^2thetatantheta`
C. `asin^2thetacostheta`
D. `asin^3thetatantheta`

Answer

To find the length of the normal at a given angle \(\theta\) on the curve defined by the parametric equations \(x = a \cos^3 \theta\) and \(y = a \sin^3 \theta\), we will follow these steps:

### Step 1: Find \(\frac{dy}{dx}\)

We start by finding \(\frac{dy}{dx}\) using the chain rule. We know that:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]

First, we need to calculate \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\).

- For \(x = a \cos^3 \theta\):
\[ \frac{dx}{d\theta} = a \cdot 3 \cos^2 \theta \cdot (-\sin \theta) = -3a \cos^2 \theta \sin \theta \]

- For \(y = a \sin^3 \theta\):
\[ \frac{dy}{d\theta} = a \cdot 3 \sin^2 \theta \cdot \cos \theta = 3a \sin^2 \theta \cos \theta \]

Now substituting these into the formula for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta \]

### Step 2: Use the formula for the length of the normal

The formula for the length of the normal at a point on a curve is given by:
\[ L = |y| \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \]

Substituting \(\frac{dy}{dx} = -\tan \theta\):
\[ L = |y| \sqrt{1 + \tan^2 \theta} \]

Using the identity \(1 + \tan^2 \theta = \sec^2 \theta\):
\[ L = |y| \cdot \sec \theta \]

### Step 3: Substitute the value of \(y\)

Now we substitute \(y = a \sin^3 \theta\):
\[ L = |a \sin^3 \theta| \cdot \sec \theta \]

Since \(a\) is a positive constant, we can drop the absolute value:
\[ L = a \sin^3 \theta \cdot \sec \theta \]

Using the identity \(\sec \theta = \frac{1}{\cos \theta}\):
\[ L = a \sin^3 \theta \cdot \frac{1}{\cos \theta} = a \frac{\sin^3 \theta}{\cos \theta} = a \sin^2 \theta \tan \theta \]

### Final Result

Thus, the length of the normal at \(\theta\) on the curve is:
\[ L = a \sin^2 \theta \tan \theta \]