The equation is a circular helix. In my understanding, if a particle moving along the helix, its direction of tangent should be changing.
For example, the tangent vector v should not be <0,1,1> if time is not 0.

So, what is the meaning of r(t)? Why it uses t<0,1,1>?

It said that r(t) is the position vector of a testing point Q one the line, but I don't understand which line it refers to.

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Example 1: Given that R(t) = (cost)i+ (sin t)j+ (t)k is a vector function whose graph is a circular helix. Find the vector equation of the %3D tangent line to the curve at point t = 0. Solution to Example 1: Notice that every line in 3-space is determined by (1) a vector which is parallel to the line, (2) a point lying on the straight line. 1) Things to do: Find a vector which is parallel to the concerned straight line. In this example, the tangent vector R'(0) is parallel to the line. By definition, we have R'(t) dF dF, dFz del del >=<- sint ,cost ,1 >. Hence dt dt dt dt R' (0) = <0,1,1 >=j+k. 2) Ihings to do: Find a point lying on the concerned line. In this example, the concerned line passes through the point R(0) = =<0,1,1 > = <1,0,0>. The vector equation of the line is given by 7(t) = r+ tu =<1,0,0>+t <0,1,1> where 7(t) is the position vector of %3D a testing point Q on the line. The parametric equations of the line are given by x = x(t) 3D y = y(t) = 0 +(1)t where z = z(t) = 0 +(1)t 1 +(0)t %3D -o <t<8. %3D The physical interpretation of vector-valued functions, their derivatives, and integrals: Let R be the position vector of a particle moving in 3-space at time t. dt dt R. dR is called the particle's velocity vector at timet, which is tangent to the curve of R. At any The vector function above el at time, the velocity vector shows the direction of motion of the moving partic le. The magnitude of is called the particle's speed. The derivative d is called the partic le's acceleration vector at time t. at SF Example 1: Given that R(t) = (cost)i+ (sin t)j+ (t)k is a vector function whose graph is a circular helix. Find the vector equation of the %3D tangent line to the curve at point t = 0. Solution to Example 1: Notice that every line in 3-space is determined by (1) a vector which is parallel to the line, (2) a point lying on the straight line. 1) Things to do: Find a vector which is parallel to the concerned straight line. In this example, the tangent vector R'(0) is parallel to the line. By definition, we have R'(t) dF dF, dFz del del >=<- sint ,cost ,1 >. Hence dt dt dt dt R' (0) = <0,1,1 >=j+k. 2) Ihings to do: Find a point lying on the concerned line. In this example, the concerned line passes through the point R(0) = =<0,1,1 > = <1,0,0>. The vector equation of the line is given by 7(t) = r+ tu =<1,0,0>+t <0,1,1> where 7(t) is the position vector of %3D a testing point Q on the line. The parametric equations of the line are given by x = x(t) 3D y = y(t) = 0 +(1)t where z = z(t) = 0 +(1)t 1 +(0)t %3D -o <t<8. %3D The physical interpretation of vector-valued functions, their derivatives, and integrals: Let R be the position vector of a particle moving in 3-space at time t. dt dt R. dR is called the particle's velocity vector at timet, which is tangent to the curve of R. At any The vector function above el at time, the velocity vector shows the direction of motion of the moving partic le. The magnitude of is called the particle's speed. The derivative d is called the partic le's acceleration vector at time t. at SF

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