consider the path r(t)=(14t,7t^2,7 ln t) defined for t > 0. find the length of the curve between the points…

Explanation:

Step1: Find the derivative of $\mathbf{r}(t)$

If $\mathbf{r}(t)=(14t,7t^{2},7\ln t)$, then $\mathbf{r}'(t)=(14,14t,\frac{7}{t})$.

Step2: Calculate the magnitude of $\mathbf{r}'(t)$

$|\mathbf{r}'(t)|=\sqrt{14^{2}+(14t)^{2}+(\frac{7}{t})^{2}}=\sqrt{196 + 196t^{2}+\frac{49}{t^{2}}}=\sqrt{49(4 + 4t^{2}+\frac{1}{t^{2}})} = 7\sqrt{(2t+\frac{1}{t})^{2}}=7(2t+\frac{1}{t})$ (since $t>0$).

Step3: Determine the values of $t$ corresponding to the given points

For the point $(14,7,0)$, when $14t = 14$, $t = 1$; when $7t^{2}=7$, $t = 1$; when $7\ln t=0$, $t = 1$. For the point $(70,175,7\ln5)$, when $14t = 70$, $t = 5$; when $7t^{2}=175$, $t = 5$; when $7\ln t=7\ln5$, $t = 5$.

Step4: Calculate the arc - length using the arc - length formula

The arc - length formula is $L=\int_{a}^{b}|\mathbf{r}'(t)|dt$. Here $a = 1$, $b = 5$, and $|\mathbf{r}'(t)|=7(2t+\frac{1}{t})$. $L=\int_{1}^{5}7(2t+\frac{1}{t})dt=7\int_{1}^{5}(2t+\frac{1}{t})dt$. $=7\left[t^{2}+\ln t\right]_{1}^{5}=7\left((5^{2}+\ln5)-(1^{2}+\ln1)\right)=7(25 + \ln5 - 1)=7(24+\ln5)=168 + 7\ln5$.

Answer:

$168 + 7\ln5$