Let `C:vecr(t)=x(t)hati+y(t)hatj+z(t)hatk` be a differentiable curve, i.e. `exists lim_(hto0) (vecr(t+h)-vecr(t))/(h) AA t`
`therefore vecr'(t)=x'(t)hati+y'(t)hatj+z'(t)hatk`
`vecr'(t)` is tangent to the curve `C` at the point `P[x(t),y(t),z(t)]` and points in the direction of increasing `t`.
The tangent vector to `vecr(t)=(2t^(2))hati+(1-t)hatj+(3t^(2)+2)hatk` at `(2,0,5)` is:

Options

A. (a) `4hati+hatj-6hatk`
B. (b) `4hati-hatj+6hatk`
C. (c) `2hati-hatj+6hatk`
D. (d) `2hati+hatj-6hatk`

Explanation

To find the tangent vector to the curve defined by \(\vec{r}(t) = 2t^2 \hat{i} + (1 - t) \hat{j} + (3t^2 + 2) \hat{k}\) at the point \((2, 0, 5)\), we will follow these steps:

### Step 1: Identify the components of the vector function
The vector function is given as:
\[ \vec{r}(t) = x(t) \hat{i} + y(t) \hat{j} + z(t) \hat{k} \]
where:
- \(x(t) = 2t^2\)
- \(y(t) = 1 - t\)
- \(z(t) = 3t^2 + 2\)

### Step 2: Find the derivatives of the components
We need to compute the derivatives of \(x(t)\), \(y(t)\), and \(z(t)\):
\[ x'(t) = \frac{d}{dt}(2t^2) = 4t \]
\[ y'(t) = \frac{d}{dt}(1 - t) = -1 \]
\[ z'(t) = \frac{d}{dt}(3t^2 + 2) = 6t \]

### Step 3: Determine the value of \(t\) at the point \((2, 0, 5)\)
We need to find \(t\) such that:
\[ x(t) = 2, \quad y(t) = 0, \quad z(t) = 5 \]
From \(x(t) = 2t^2\):
\[ 2t^2 = 2 \implies t^2 = 1 \implies t = 1 \quad (\text{since } t \text{ must be non-negative}) \]

From \(y(t) = 1 - t\):
\[ 1 - t = 0 \implies t = 1 \]

From \(z(t) = 3t^2 + 2\):
\[ 3t^2 + 2 = 5 \implies 3t^2 = 3 \implies t^2 = 1 \implies t = 1 \]

Thus, \(t = 1\) satisfies all three equations.

### Step 4: Evaluate the derivatives at \(t = 1\)
Now we substitute \(t = 1\) into the derivatives:
\[ x'(1) = 4(1) = 4 \]
\[ y'(1) = -1 \]
\[ z'(1) = 6(1) = 6 \]

### Step 5: Form the tangent vector
The tangent vector \(\vec{r}'(t)\) at \(t = 1\) is given by:
\[ \vec{r}'(1) = x'(1) \hat{i} + y'(1) \hat{j} + z'(1) \hat{k} = 4 \hat{i} - 1 \hat{j} + 6 \hat{k} \]

### Final Answer
Thus, the tangent vector at the point \((2, 0, 5)\) is:
\[ \vec{r}'(1) = 4 \hat{i} - \hat{j} + 6 \hat{k} \]