For the functions defined parametrically by the equations
`f(t)=x={{:(2t+t^(2)sin.(1)/(t),,,tne0),(0,,,t=0):}` and
`g(t)=y={{:((1)/(t)"sint"^(2),t ne0),(0,t =0):}`

Options

A. equation of tangent at t = 0 is `x-2y=0`
B. equation of normal at t = 0 is `2x+y=0`
C. tangent does not exist at t = 0
D. normal does not exist at `t = 0`

Explanation

To find \(\frac{dy}{dx}\) for the given parametric equations, we will follow these steps:

### Step 1: Define the Parametric Equations
The functions are given as:
\[ f(t) = \begin{cases} 2t + t^2 \sin\left(\frac{1}{t}\right), & t \neq 0 \\ 0, & t = 0 \end{cases} \]
\[ g(t) = \begin{cases} \frac{\sin(t^2)}{t}, & t \neq 0 \\ 0, & t = 0 \end{cases} \]

### Step 2: Find \(g'(t)\)
To find \(\frac{dy}{dx}\), we need to compute \(g'(t)\) first. We will use the limit definition of the derivative:
\[ g'(0) = \lim_{t \to 0} \frac{g(t) - g(0)}{t - 0} = \lim_{t \to 0} \frac{\frac{\sin(t^2)}{t} - 0}{t} \]
This simplifies to:
\[ g'(0) = \lim_{t \to 0} \frac{\sin(t^2)}{t^2} = 1 \]
(using the fact that \(\lim_{u \to 0} \frac{\sin(u)}{u} = 1\) where \(u = t^2\)).

### Step 3: Find \(f'(t)\)
Next, we compute \(f'(t)\):
\[ f'(0) = \lim_{t \to 0} \frac{f(t) - f(0)}{t - 0} = \lim_{t \to 0} \frac{2t + t^2 \sin\left(\frac{1}{t}\right) - 0}{t} \]
This simplifies to:
\[ f'(0) = \lim_{t \to 0} \left(2 + t \sin\left(\frac{1}{t}\right)\right) \]
As \(t \to 0\), \(t \sin\left(\frac{1}{t}\right)\) approaches \(0\), so:
\[ f'(0) = 2 + 0 = 2 \]

### Step 4: Compute \(\frac{dy}{dx}\)
Now we can find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{g'(t)}{f'(t)} = \frac{1}{2} \]

### Step 5: Find the Equation of the Tangent Line
The slope of the tangent line at \(t = 0\) is \(\frac{1}{2}\). The equation of the tangent line at the point \((f(0), g(0)) = (0, 0)\) is given by:
\[ y - 0 = \frac{1}{2}(x - 0) \implies y = \frac{1}{2}x \]

### Step 6: Find the Equation of the Normal Line
The slope of the normal line is the negative reciprocal of the slope of the tangent line:
\[ \text{slope of normal} = -\frac{1}{\frac{1}{2}} = -2 \]
Thus, the equation of the normal line is:
\[ y - 0 = -2(x - 0) \implies y = -2x \]

### Final Result
The equations of the tangent and normal lines at the point \((0, 0)\) are:
- Tangent: \(y = \frac{1}{2}x\)
- Normal: \(y = -2x\)

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