Find `dy/dx`, if x and y are connected parametrically by the equations, given below without eliminating the parameter:
`x=e^(t)cost,y=e^(t)sint" at "t=pi/2`.

Answer

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

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

1. Differentiate \(x = e^t \cos t\):
\[ \frac{dx}{dt} = \frac{d}{dt}(e^t \cos t) = e^t \cos t + e^t (-\sin t) = e^t (\cos t - \sin t) \]

2. Differentiate \(y = e^t \sin t\):
\[ \frac{dy}{dt} = \frac{d}{dt}(e^t \sin t) = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t) \]

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

Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{e^t (\sin t + \cos t)}{e^t (\cos t - \sin t)} \]

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

The \(e^t\) terms cancel out:
\[ \frac{dy}{dx} = \frac{\sin t + \cos t}{\cos t - \sin t} \]

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

Now we substitute \(t = \frac{\pi}{2}\):
- \(\sin\left(\frac{\pi}{2}\right) = 1\)
- \(\cos\left(\frac{\pi}{2}\right) = 0\)

Thus,
\[ \frac{dy}{dx} = \frac{1 + 0}{0 - 1} = \frac{1}{-1} = -1 \]

### Final Answer
\[ \frac{dy}{dx} \text{ at } t = \frac{\pi}{2} \text{ is } -1 \]

---