Question

calculate r(t) and t(t), where r(t) = <6t - 1, -(6 + t), 7 + 6t>. r(t) = <, , >. t(t) = <, , >. submit answer next item answers # your answer score 01 - - / 0.5 02 - - / 0.5 03 - - / 0.5 04 - - / 0.5

Explanation:

Step1: Differentiate each component of r(t)

If \(r(t)=\langle x(t),y(t),z(t)
angle=\langle6t - 1,-(6 + t),7+6t
angle\), 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, \(r'(t)=\langle6,-1,6
angle\).

Step2: Calculate the magnitude of r'(t)

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

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

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

Answer:

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