Write the equation of a tangent to the curve `x=t, y=t^2 and z=t^3` at its point `M(1, 1, 1): (t=1)`.

Answer

To find the equation of the tangent to the curve defined by the parametric equations \( x = t \), \( y = t^2 \), and \( z = t^3 \) at the point \( M(1, 1, 1) \) where \( t = 1 \), we will follow these steps:

### Step 1: Differentiate the parametric equations
We need to differentiate the equations with respect to \( t \):

- \( \frac{dx}{dt} = 1 \)
- \( \frac{dy}{dt} = 2t \)
- \( \frac{dz}{dt} = 3t^2 \)

### Step 2: Evaluate the derivatives at \( t = 1 \)
Now, we will evaluate these derivatives at \( t = 1 \):

- \( \frac{dx}{dt} \bigg|_{t=1} = 1 \)
- \( \frac{dy}{dt} \bigg|_{t=1} = 2(1) = 2 \)
- \( \frac{dz}{dt} \bigg|_{t=1} = 3(1^2) = 3 \)

### Step 3: Write the direction ratios of the tangent
The direction ratios of the tangent line at the point \( M(1, 1, 1) \) are given by the derivatives we just calculated:

- Direction ratios: \( (1, 2, 3) \)

### Step 4: Use the point-direction form of the line
The equation of a line in space can be expressed in the parametric form as:

\[ \begin{align*} x &= x_0 + at \\ y &= y_0 + bt \\ z &= z_0 + ct \end{align*} \]

Where \( (x_0, y_0, z_0) \) is the point on the line and \( (a, b, c) \) are the direction ratios. Here, \( (x_0, y_0, z_0) = (1, 1, 1) \) and \( (a, b, c) = (1, 2, 3) \).

Thus, the parametric equations of the tangent line are:

\[ \begin{align*} x &= 1 + 1t \\ y &= 1 + 2t \\ z &= 1 + 3t \end{align*} \]

### Step 5: Write the symmetric form of the equation
To express the tangent line in symmetric form, we can eliminate \( t \):

\[ \frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 1}{3} \]

### Final Answer
The equation of the tangent to the curve at the point \( M(1, 1, 1) \) is given by:

\[ \frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 1}{3} \]