Tangent Line at t=0 for Parametric Curve (8sin(5t)-3, 8cos(5t)+3, 3t-4)

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SUMMARY

The tangent line at t=0 for the parametric curve defined by the equations x=8sin(5t)-3, y=8cos(5t)+3, and z=3t-4 is derived using the derivatives of the parametric equations. The derivatives at t=0 yield the values x'(0)=40, y'(0)=0, and z'(0)=3, resulting in the displacement vector (40, 0, 3). The parametric equation for the tangent line can be expressed as r(t) = (x(0), y(0), z(0)) + h(40, 0, 3), where (x(0), y(0), z(0)) is the position vector at t=0.

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alchal
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Homework Statement



If the parametric equations of a curve C are (x= 8sin(5 t) -3, y=8cos(5 t) + 3, z=3 t -4),
what are the parametric equations of the tangent line to this curve at t=0 ?

The Attempt at a Solution



x(t)=8sin(5t)-3
y(t)=8cos(5t)+3
z(t)=3t-4

x'(t)=40cos(5t)
y'(t)=-40sin(5t)
z'(t)=3

x'(0)=40
y'(0)=0
z'(0)=3

not even sure if I am doing this right...
 
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You're doing alright so far. A parametric equation for the tangent line will take the form a + hb, where a is a position vector for the point on your original curve at t = 0, h is a scalar, namely the parameter for this new equation describing the tangent line, and b is a displacement vector tangent to your original curve at t = 0, for example the derivative you found: (40,0,3).
 

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