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Sagot :
The tangent vector to the curve at t = n is given by the derivative of the parametric equations: [tex]\frac{dx}{dt} = nt \ and\ \frac{dy}{dt} = n't^2[/tex]
To find the principal unit normal vector to a curve at a specified value of the parameter, you will need to do the following:
- Find the tangent vector to the curve at the specified value of the parameter. The tangent vector is a vector that is tangent to the curve at the point in question and points in the direction of the curve at that point.
- Find the normal vector to the curve at the specified value of the parameter. The normal vector is a vector that is perpendicular to the tangent vector and points in a direction that is normal (perpendicular) to the curve at the point in question.
- Normalize the normal vector to find the principal unit normal vector. Normalization is the process of dividing a vector by its magnitude (length) to obtain a vector of length 1, which is called a unit vector.
Here is an example to illustrate this process:
Suppose we have a curve given by the parametric equations [tex]x = t^2[/tex] and y = t^3. The tangent vector to the curve at t = 2 is given by the derivative of the parametric equations:
[tex]\frac{dx}{dt} = 2t \ and\ \frac{dy}{dt} = 3t^2[/tex]
Evaluating these at t = 2, we find that the tangent vector is (4,12).
The normal vector to the curve at t = 2 is perpendicular to the tangent vector, so we can find it by rotating the tangent vector 90 degrees counterclockwise. This can be done by taking the negative of the second component and swapping the components: (-12,4).
To normalize the normal vector, we divide it by its magnitude:
[tex](-12,4)\sqrt((-12)^2 + 4^2) = (-12\sqrt164,4\sqrt164) = (-0.73,0.24)[/tex]
This is the principal unit normal vector to the curve at t = 2.
To learn more about tangent, visit:
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