for the curve given by $r(t)=langle 1t,e^{5t},e^{-5t}\rangle$, find the derivative $r(t)=langle 1,5e^{5t}…

Explanation:

Step1: Recall curvature formula

The formula for the curvature $\kappa(t)$ of a vector - valued function $r(t)=\langle x(t),y(t),z(t)\rangle$ is $\kappa(t)=\frac{\left\lVert r^{\prime}(t)\times r^{\prime\prime}(t)\right\rVert}{\left\lVert r^{\prime}(t)\right\rVert^{3}}$. First, we have $r(t)=\langle t,e^{5t},e^{- 5t}\rangle$, $r^{\prime}(t)=\langle1,5e^{5t},-5e^{-5t}\rangle$ and $r^{\prime\prime}(t)=\langle0,25e^{5t},25e^{-5t}\rangle$.

Step2: Calculate the cross - product $r^{\prime}(t)\times r^{\prime\prime}(t)$

The cross - product of two vectors $\vec{a}=\langle a_1,a_2,a_3\rangle$ and $\vec{b}=\langle b_1,b_2,b_3\rangle$ is $\vec{a}\times\vec{b}=\langle a_2b_3 - a_3b_2,a_3b_1 - a_1b_3,a_1b_2 - a_2b_1\rangle$. So, $r^{\prime}(t)\times r^{\prime\prime}(t)=\left\langle(5e^{5t})\times(25e^{-5t})-(-5e^{-5t})\times(25e^{5t}),( - 5e^{-5t})\times0 - 1\times(25e^{-5t}),1\times(25e^{5t})-5e^{5t}\times0\right\rangle=\langle250, - 25e^{-5t},25e^{5t}\rangle$.

Step3: Evaluate at $t = 0$

When $t = 0$:

• $r^{\prime}(0)=\langle1,5,-5\rangle$, so $\left\lVert r^{\prime}(0)\right\rVert=\sqrt{1^{2}+5^{2}+(-5)^{2}}=\sqrt{1 + 25+25}=\sqrt{51}$.
• $r^{\prime}(0)\times r^{\prime\prime}(0)=\langle250,-25,25\rangle$, so $\left\lVert r^{\prime}(0)\times r^{\prime\prime}(0)\right\rVert=\sqrt{250^{2}+(-25)^{2}+25^{2}}=\sqrt{62500 + 625+625}=\sqrt{63750}=25\sqrt{102}$.

Step4: Calculate the curvature at $t = 0$

$\kappa(0)=\frac{\left\lVert r^{\prime}(0)\times r^{\prime\prime}(0)\right\rVert}{\left\lVert r^{\prime}(0)\right\rVert^{3}}=\frac{25\sqrt{102}}{(\sqrt{51})^{3}}=\frac{25\sqrt{102}}{51\sqrt{51}}=\frac{25\sqrt{2\times51}}{51\sqrt{51}}=\frac{25\sqrt{2}}{51}$.

Answer:

$\frac{25\sqrt{2}}{51}$