A curve is difined parametrically by `x=e^(sqrtt),y=3t-log_(e)(t^(2)),` where t is a parameter. Then the equation of the tangent line drawn to the curve at t = 1 is

Options

A. `y=(2)/(e)x+1`
B. `y=(2)/(e)x-1`
C. `y=(e)/(2)x+1`
D. `y=(e)/(2)x-1`

Answer

To find the equation of the tangent line to the parametric curve defined by \( x = e^{\sqrt{t}} \) and \( y = 3t - \log_e(t^2) \) at \( t = 1 \), we will follow these steps:

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

1. **Differentiate \( x \)**:
\[ x = e^{\sqrt{t}} \]
Using the chain rule:
\[ \frac{dx}{dt} = e^{\sqrt{t}} \cdot \frac{d}{dt}(\sqrt{t}) = e^{\sqrt{t}} \cdot \frac{1}{2\sqrt{t}} \]

2. **Differentiate \( y \)**:
\[ y = 3t - \log_e(t^2) \]
Simplifying \( \log_e(t^2) \) gives \( 2\log_e(t) \):
\[ y = 3t - 2\log_e(t) \]
Now, differentiating:
\[ \frac{dy}{dt} = 3 - 2 \cdot \frac{1}{t} = 3 - \frac{2}{t} \]

### Step 2: Evaluate the derivatives at \( t = 1 \)

1. **Evaluate \( \frac{dx}{dt} \) at \( t = 1 \)**:
\[ \frac{dx}{dt} \bigg|_{t=1} = e^{\sqrt{1}} \cdot \frac{1}{2\sqrt{1}} = e \cdot \frac{1}{2} = \frac{e}{2} \]

2. **Evaluate \( \frac{dy}{dt} \) at \( t = 1 \)**:
\[ \frac{dy}{dt} \bigg|_{t=1} = 3 - \frac{2}{1} = 3 - 2 = 1 \]

### Step 3: Find the slope of the tangent line

The slope of the tangent line \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the values we found:
\[ \frac{dy}{dx} \bigg|_{t=1} = \frac{1}{\frac{e}{2}} = \frac{2}{e} \]

### Step 4: Find the coordinates of the point on the curve at \( t = 1 \)

1. **Calculate \( x \) at \( t = 1 \)**:
\[ x = e^{\sqrt{1}} = e \]

2. **Calculate \( y \) at \( t = 1 \)**:
\[ y = 3 \cdot 1 - \log_e(1^2) = 3 - 0 = 3 \]

Thus, the point on the curve at \( t = 1 \) is \( (e, 3) \).

### Step 5: Write the equation of the tangent line

Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \( m = \frac{2}{e} \), \( (x_1, y_1) = (e, 3) \):
\[ y - 3 = \frac{2}{e}(x - e) \]
Rearranging gives:
\[ y - 3 = \frac{2}{e}x - 2 \]
\[ y = \frac{2}{e}x + 1 \]

### Final Answer

The equation of the tangent line is:
\[ y = \frac{2}{e}x + 1 \]