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We are asked to find the parametric equations for the tangent line to the curve given by:

\[
x = \sqrt{t^2 + 3}, \quad y = \ln(t^2 + 3), \quad z = t
\]

at the point \((2, \ln 4, 1)\).

### Step 1: Verify the value of \( t \) at the point \((2, \ln 4, 1)\)
The point \((x, y, z)\) is given as \((2, \ln 4, 1)\). We know \( z = t \), so:
\[
t = 1.
\]

### Step 2: Compute the tangent vector
To find the tangent vector, we need to compute the derivatives of \(x(t)\), \(y(t)\), and \(z(t)\) with respect to \(t\).

1. **Derivative of \(x(t)\):**
   \[
   x(t) = \sqrt{t^2 + 3}.
   \]
   Using the chain rule:
   \[
   \frac{dx}{dt} = \frac{1}{2}(t^2 + 3)^{-\frac{1}{2}} \cdot 2t = \frac{t}{\sqrt{t^2 + 3}}.
   \]
   At \( t = 1 \):
   \[
   \frac{dx}{dt}\bigg|_{t=1} = \frac{1}{\sqrt{1^2 + 3}} = \frac{1}{2}.
   \]

2. **Derivative of \(y(t)\):**
   \[
   y(t) = \ln(t^2 + 3).
   \]
   Using the chain rule:
   \[
   \frac{dy}{dt} = \frac{1}{t^2 + 3} \cdot 2t = \frac{2t}{t^2 + 3}.
   \]
   At \( t = 1 \):
   \[
   \frac{dy}{dt}\bigg|_{t=1} = \frac{2 \cdot 1}{1^2 + 3} = \frac{2}{4} = \frac{1}{2}.
   \]

3. **Derivative of \(z(t)\):**
   \[
   z(t) = t \quad \Rightarrow \quad \frac{dz}{dt} = 1.
   \]

Thus, the tangent vector at \( t = 1 \) is:
\[
\left( \frac{1}{2}, \frac{1}{2}, 1 \right).
\]

### Step 3: Parametric equations of the tangent line
The parametric equations for the tangent line can be written in the form:
\[
x = x_0 + v_x t, \quad y = y_0 + v_y t, \quad z = z_0 + v_z t,
\]
where \((x_0, y_0, z_0)\) is the point on the curve and \((v_x, v_y, v_z)\) is the tangent vector.

Using the point \((2, \ln 4, 1)\) and the tangent vector \(\left( \frac{1}{2}, \frac{1}{2}, 1 \right)\), we get the parametric equations for the tangent line:
\[
x = 2 + \frac{1}{2} t, \quad y = \ln 4 + \frac{1}{2} t, \quad z = 1 + t.
\]

These are the parametric equations of the tangent line.

Would you like more details on any of these steps, or do you have further questions? Here are some related problems you might find interesting:

1. How do you find the normal vector to a curve in space at a given point?
2. Can parametric curves have tangent vectors with zero magnitude?
3. How do you find the arc length of a parametric curve?
4. What happens when the tangent vector of a parametric curve becomes undefined?
5. How do you find the osculating plane of a parametric curve at a point?

**Tip:** When working with parametric curves, it's useful to keep track of both the derivatives and the geometric meaning of those derivatives (e.g., tangent vectors, normal vectors).

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point: x = sqrt(t^2 + 3), y = ln(t^2 + 3), z = t at the point (2, ln 4, 1).

Learn how to find the parametric equations for the tangent line to a curve at the point (2, ln 4, 1). This problem uses calculus concepts, including the chain rule and derivatives of logarithmic and square root functions. Suitable for advanced high school students.

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