If `x=sint-tcost" and "y=tsint+cos t`. then what is `(dy)/(dx)` at point `t=(pi)/(2)` ?

Options

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

Answer

To find \(\frac{dy}{dx}\) at the point \(t = \frac{\pi}{2}\), we start with the given equations:

\[ x = \sin t - t \cos t \]
\[ y = t \sin t + \cos t \]

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

We will differentiate \(x\) and \(y\) with respect to \(t\).

**For \(x\):**
\[ \frac{dx}{dt} = \frac{d}{dt}(\sin t - t \cos t) \]
Using the product rule on \(t \cos t\):
\[ \frac{dx}{dt} = \cos t - \left( \cos t + t (-\sin t) \right) = \cos t - \cos t + t \sin t = t \sin t \]

**For \(y\):**
\[ \frac{dy}{dt} = \frac{d}{dt}(t \sin t + \cos t) \]
Again, using the product rule on \(t \sin t\):
\[ \frac{dy}{dt} = \sin t + t \cos t - \sin t = t \cos t \]

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

Using the chain rule, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{t \cos t}{t \sin t} = \frac{\cos t}{\sin t} = \cot t \]

### Step 3: Evaluate \(\frac{dy}{dx}\) at \(t = \frac{\pi}{2}\)

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

### Final Answer
\[ \frac{dy}{dx} \text{ at } t = \frac{\pi}{2} \text{ is } 0. \]

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