Answer:
x = 5 + 10t
y = -1 + 3t
z = 3 + 9t²
Explanation:
To find the parametric equation for the line that is tangent to the curve r(t) = (5t², 3t - 4, 3t³) at t = t₀ = 1, we can follow these steps:
Step 1: Find the derivative of the curve r(t) with respect to t. This will give us the velocity vector of the curve, which represents the direction of the tangent line at any point on the curve.
Taking the derivative of r(t) with respect to t, we get:
r'(t) = (10t, 3, 9t²)
Step 2: Substitute t = t₀ = 1 into the velocity vector r'(t) to find the direction of the tangent line at t = 1.
Substituting t = 1 into r'(t), we get:
r'(1) = (10, 3, 9)
Step 3: Write the parametric equation for the line that is tangent to the curve at t = 1.
Using the point-direction form of a line equation, the parametric equation can be written as:
x = 5t₀² + at
y = 3t₀ - 4 + bt
z = 3t₀³ + ct
where a, b, and c are the components of the direction vector r'(1).
Substituting the values of t₀ = 1 and r'(1) into the parametric equation, we get:
x = 5(1)² + 10t = 5 + 10t
y = 3(1) - 4 + 3t = -1 + 3t
z = 3(1)³ + 9t² = 3 + 9t²
Therefore, the parametric equation for the line tangent to the curve r(t) = (5t², 3t - 4, 3t³) at t = 1 is:
x = 5 + 10t
y = -1 + 3t
z = 3 + 9t²