If `x=sin t-t cos t and y = t sin t +cos t,` then what is `(dy)/(dx)`
at point `t=(pi)/(2)?`

Options

A. 0
B. `(pi)/(2)`
C. `-(pi)/(2)`
D. 1

Explanation

To find \(\frac{dy}{dx}\) at the point \(t = \frac{\pi}{2}\) for the given parametric equations \(x = \sin t - t \cos t\) and \(y = t \sin t + \cos t\), we will follow these steps:

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

Given:
\[ y = t \sin t + \cos t \]

To find \(\frac{dy}{dt}\), we differentiate \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = \frac{d}{dt}(t \sin t) + \frac{d}{dt}(\cos t) \]
Using the product rule for \(t \sin t\):
\[ \frac{d}{dt}(t \sin t) = \sin t + t \cos t \]
And the derivative of \(\cos t\) is \(-\sin t\):
\[ \frac{dy}{dt} = \sin t + t \cos t - \sin t = t \cos t \]

### Step 2: Find \(\frac{dx}{dt}\)

Given:
\[ x = \sin t - t \cos t \]

To find \(\frac{dx}{dt}\), we differentiate \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = \frac{d}{dt}(\sin t) - \frac{d}{dt}(t \cos t) \]
Using the product rule for \(t \cos t\):
\[ \frac{d}{dt}(t \cos t) = \cos t - t \sin t \]
Thus,
\[ \frac{dx}{dt} = \cos t - (\cos t - t \sin t) = t \sin t \]

### Step 3: Find \(\frac{dy}{dx}\)

Now, we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{t \cos t}{t \sin t} \]
Simplifying this gives:
\[ \frac{dy}{dx} = \frac{\cos t}{\sin t} = \cot t \]

### Step 4: Evaluate \(\frac{dy}{dx}\) at \(t = \frac{\pi}{2}\)

Now we substitute \(t = \frac{\pi}{2}\):
\[ \frac{dy}{dx} \bigg|_{t = \frac{\pi}{2}} = \cot\left(\frac{\pi}{2}\right) \]
Since \(\cot\left(\frac{\pi}{2}\right) = 0\):
\[ \frac{dy}{dx} \bigg|_{t = \frac{\pi}{2}} = 0 \]

### Final Answer
Thus, the value of \(\frac{dy}{dx}\) at \(t = \frac{\pi}{2}\) is:
\[ \boxed{0} \]