if ` x= f (t) cos t- f'(t) sin t and y= f (t) sin t + f'(t ) cos t, then ((dx)/(dt))^2 +((dy)/(dt))^2`=

Options

A. `f-f''`
B. ` (f-f' ' )^2`
C. `(f+f' ' )^2`
D. None

Answer

To solve the problem, we need to find the expression for \(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\) given:

\[ x = f(t) \cos t - f'(t) \sin t \]
\[ y = f(t) \sin t + f'(t) \cos t \]

### Step 1: Differentiate \(x\) with respect to \(t\)

Using the product rule for differentiation, we differentiate \(x\):

\[ \frac{dx}{dt} = \frac{d}{dt}[f(t) \cos t] - \frac{d}{dt}[f'(t) \sin t] \]

Applying the product rule:

1. For \(f(t) \cos t\):
\[ \frac{d}{dt}[f(t) \cos t] = f'(t) \cos t - f(t) \sin t \]

2. For \(f'(t) \sin t\):
\[ \frac{d}{dt}[f'(t) \sin t] = f''(t) \sin t + f'(t) \cos t \]

Combining these results, we have:

\[ \frac{dx}{dt} = (f'(t) \cos t - f(t) \sin t) - (f''(t) \sin t + f'(t) \cos t) \]

Simplifying:

\[ \frac{dx}{dt} = -f(t) \sin t - f''(t) \sin t \]

Factoring out \(\sin t\):

\[ \frac{dx}{dt} = -\sin t (f(t) + f''(t)) \]

### Step 2: Differentiate \(y\) with respect to \(t\)

Now, we differentiate \(y\):

\[ \frac{dy}{dt} = \frac{d}{dt}[f(t) \sin t] + \frac{d}{dt}[f'(t) \cos t] \]

Applying the product rule:

1. For \(f(t) \sin t\):
\[ \frac{d}{dt}[f(t) \sin t] = f'(t) \sin t + f(t) \cos t \]

2. For \(f'(t) \cos t\):
\[ \frac{d}{dt}[f'(t) \cos t] = f''(t) \cos t - f'(t) \sin t \]

Combining these results, we have:

\[ \frac{dy}{dt} = (f'(t) \sin t + f(t) \cos t) + (f''(t) \cos t - f'(t) \sin t) \]

Simplifying:

\[ \frac{dy}{dt} = f(t) \cos t + f''(t) \cos t \]

Factoring out \(\cos t\):

\[ \frac{dy}{dt} = \cos t (f(t) + f''(t)) \]

### Step 3: Calculate \(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\)

Now, we calculate:

\[ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \]

Substituting the expressions we found:

\[ \left(-\sin t (f(t) + f''(t))\right)^2 + \left(\cos t (f(t) + f''(t))\right)^2 \]

This simplifies to:

\[ \sin^2 t (f(t) + f''(t))^2 + \cos^2 t (f(t) + f''(t))^2 \]

Factoring out \((f(t) + f''(t))^2\):

\[ = (f(t) + f''(t))^2 (\sin^2 t + \cos^2 t) \]

Using the Pythagorean identity \(\sin^2 t + \cos^2 t = 1\):

\[ = (f(t) + f''(t))^2 \]

### Final Result

Thus, we conclude:

\[ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = (f(t) + f''(t))^2 \]