If `x=e^(t)cost` and `y=e^(t)sint`, then what is `(dx)/(dy)` at t=0 equal to?

Options

A. 0
B. 1
C. 2e
D. `-1`

Answer

To find \(\frac{dx}{dy}\) at \(t=0\) given the parametric equations \(x = e^t \cos t\) and \(y = e^t \sin t\), we follow these steps:

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

We start with the equation for \(x\):
\[ x = e^t \cos t \]
Using the product rule, we differentiate \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = \frac{d}{dt}(e^t) \cdot \cos t + e^t \cdot \frac{d}{dt}(\cos t) \]
Calculating the derivatives:
\[ \frac{d}{dt}(e^t) = e^t \quad \text{and} \quad \frac{d}{dt}(\cos t) = -\sin t \]
Substituting these into the equation gives:
\[ \frac{dx}{dt} = e^t \cos t - e^t \sin t = e^t (\cos t - \sin t) \]

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

Next, we differentiate \(y\):
\[ y = e^t \sin t \]
Again using the product rule:
\[ \frac{dy}{dt} = \frac{d}{dt}(e^t) \cdot \sin t + e^t \cdot \frac{d}{dt}(\sin t) \]
Calculating the derivatives:
\[ \frac{d}{dt}(e^t) = e^t \quad \text{and} \quad \frac{d}{dt}(\sin t) = \cos t \]
Substituting these into the equation gives:
\[ \frac{dy}{dt} = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t) \]

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

Now we can find \(\frac{dx}{dy}\) using the chain rule:
\[ \frac{dx}{dy} = \frac{dx/dt}{dy/dt} \]
Substituting the derivatives we found:
\[ \frac{dx}{dy} = \frac{e^t (\cos t - \sin t)}{e^t (\sin t + \cos t)} \]
The \(e^t\) terms cancel out:
\[ \frac{dx}{dy} = \frac{\cos t - \sin t}{\sin t + \cos t} \]

### Step 4: Evaluate \(\frac{dx}{dy}\) at \(t=0\)

Now we evaluate \(\frac{dx}{dy}\) at \(t=0\):
\[ \frac{dx}{dy} \bigg|_{t=0} = \frac{\cos(0) - \sin(0)}{\sin(0) + \cos(0)} = \frac{1 - 0}{0 + 1} = \frac{1}{1} = 1 \]

### Final Answer
Thus, \(\frac{dx}{dy}\) at \(t=0\) is equal to \(1\).

---