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Consider the following vector function. r(t)=⟨2t,1/2t²,t²⟩

Find the unit tangent and unit normal vectors T(t) and N(t)

User Miasha
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The unit tangent and unit normal vectors, T(t) and N(t), of the vector function r(t) = ⟨2t, 1/2t², t²⟩ can be found by normalizing the derivative of the function with respect to t. the unit tangent vector T(t) is ⟨2, t, 2t⟩ / √(5t² + 4), and the unit normal vector N(t) is ⟨0, 1, 2⟩ / √5.

To find the unit tangent vector T(t), we differentiate the vector function r(t) with respect to t:

r'(t) = ⟨2, t, 2t⟩.

Next, we normalize the derivative vector to obtain the unit tangent vector:

T(t) = r'(t) / ||r'(t)||,

where ||r'(t)|| denotes the magnitude of r'(t). To find the magnitude, we calculate:

||r'(t)|| = √(2² + t² + (2t)²) = √(4 + t² + 4t²) = √(5t² + 4).

Thus, the unit tangent vector T(t) is:

T(t) = ⟨2, t, 2t⟩ / √(5t² + 4).

To find the unit normal vector N(t), we differentiate T(t) with respect to and normalize the resulting vector:

N(t) = T'(t) / ||T'(t)||.

Differentiating T(t), we get:

T'(t) = ⟨0, 1, 2⟩ / √(5t² + 4).

Normalizing T'(t), we have:

N(t) = ⟨0, 1, 2⟩ / ||⟨0, 1, 2⟩|| = ⟨0, 1, 2⟩ / √(1² + 2²) = ⟨0, 1, 2⟩ / √5.

Therefore, the unit tangent vector T(t) is ⟨2, t, 2t⟩ / √(5t² + 4), and the unit normal vector N(t) is ⟨0, 1, 2⟩ / √5.

User Oliver Friedrich
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