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Given the space curve r(t) = 3sin t i + 4t j + 3cos t k.

a) Find the curvature of r(t).
b) Find the equation of the normal plane α to the curve at point
P(0,0,3).

User Jackeline
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1 Answer

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Answer and Step-by-step explanation:

a) To find the curvature of the space curve r(t), we need to find the magnitude of the curvature vector. The curvature vector can be calculated using the formula:

Curvature = |(dT/dt)| / ||r'(t)||

where T is the unit tangent vector and r'(t) is the derivative of r(t).

First, we need to find the derivative of r(t):

r'(t) = (3cos t)i + 4j - (3sin t)k

Next, we find the magnitude of the derivative:

||r'(t)|| = √[(3cos t)² + 4² + (-3sin t)²]

= √[9cos² t + 16 + 9sin² t]

= √(25)

= 5

Now, we need to find the unit tangent vector T:

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

= [(3cos t)i + 4j - (3sin t)k] / 5

Finally, we find the magnitude of the derivative of the unit tangent vector:

|(dT/dt)| = |[-3sin t)i + 0j - (3cos t)k] / 5|

= √[(-3sin t)² + 0 + (3cos t)²] / 5

= √(9sin² t + 9cos² t) / 5

= √(9) / 5

= 3 / 5

Now, we can calculate the curvature:

Curvature = |(dT/dt)| / ||r'(t)||

= (3 / 5) / 5

= 3 / 25

Therefore, the curvature of the space curve r(t) is 3/25.

b) To find the equation of the normal plane α to the curve at point P(0,0,3), we need to find the normal vector to the plane. The normal vector is given by the derivative of the unit tangent vector dT/dt evaluated at the point P.

First, we find the derivative of T:

(dT/dt) = [-3sin t)i + 0j - (3cos t)k]

Next, we substitute t = 0 into the derivative:

(dT/dt) = [-3sin 0)i + 0j - (3cos 0)k]

= [0i + 0j - (-3)k]

= [0i + 0j + 3k]

= 3k

Since we are evaluating at point P(0,0,3), the normal vector is parallel to the z-axis.

Therefore, the equation of the normal plane α to the curve at point P(0,0,3) is of the form x = 0, y = 0, and z is free.

User Roaders
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