calculate r(t) and t(t), where r(t) = <6t - 1, -(6 + t), 7 + 6t>. r(t) = <, , >. t(t) = <, , >. submit…

Explanation:

Step1: Differentiate each component of r(t)

If (r(t)=\langle x(t),y(t),z(t)\rangle=\langle6t - 1,-(6 + t),7+6t\rangle), then using the power - rule ((ax + b)'=a) for (a) and (b) constants. (x(t)=6t - 1), so (x'(t)=6); (y(t)=-(6 + t)=-6 - t), so (y'(t)=-1); (z(t)=7 + 6t), so (z'(t)=6). Thus, (r'(t)=\langle6,-1,6\rangle).

Step2: Calculate the magnitude of r'(t)

The magnitude of a vector (v=\langle a,b,c\rangle) is (\vert v\vert=\sqrt{a^{2}+b^{2}+c^{2}}). For (r'(t)=\langle6,-1,6\rangle), (\vert r'(t)\vert=\sqrt{6^{2}+(-1)^{2}+6^{2}}=\sqrt{36 + 1+36}=\sqrt{73}).

Step3: Calculate the unit - tangent vector T(t)

The unit - tangent vector (T(t)=\frac{r'(t)}{\vert r'(t)\vert}). Since (r'(t)=\langle6,-1,6\rangle) and (\vert r'(t)\vert=\sqrt{73}), then (T(t)=\langle\frac{6}{\sqrt{73}},-\frac{1}{\sqrt{73}},\frac{6}{\sqrt{73}}\rangle).

Answer:

(r'(t)=\langle6,-1,6\rangle) (T(t)=\langle\frac{6}{\sqrt{73}},-\frac{1}{\sqrt{73}},\frac{6}{\sqrt{73}}\rangle)