What is the arc length of f(t)=(lnt,5-lnt)  over t in [3,4] ?

Arc length $l = \sqrt{2} \cdot \left(\ln 4 - \ln 3\right)$

$l = 0.406844$

Explanation:

$l = \int \sqrt{{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

Solve for $\frac{\mathrm{dx}}{\mathrm{dt}}$ given $x = \ln t$

$\frac{\mathrm{dx}}{\mathrm{dt}} = \left(\frac{1}{t}\right)$

Solve for $\frac{\mathrm{dy}}{\mathrm{dt}}$ given $y = 5 - \ln t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = - 1 \cdot \frac{1}{t}$

Solve for the length $l$

$l = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

$a = 3$ and $b = 4$

$l = {\int}_{3}^{4} \sqrt{{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

$l = {\int}_{3}^{4} \sqrt{{\left(- \frac{1}{t}\right)}^{2} + {\left(\frac{1}{t}\right)}^{2}} \mathrm{dt}$

$l = {\int}_{3}^{4} \sqrt{\frac{2}{t} ^ 2} \mathrm{dt}$

$l = {\int}_{3}^{4} \frac{\sqrt{2}}{t} \mathrm{dt}$

$l = \sqrt{2} \cdot \left(\ln 4 - \ln 3\right)$

$l = 0.406844$

God bless....I hope the explanation is useful.
The text contains a mathematical question about finding the arc length of a parametric function, and it provides a detailed solution with the final answer.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
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- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
What is the arc length of f(t)=(lnt,5-lnt)  over t in [3,4] ?

Arc length $l = \sqrt{2} \cdot \left(\ln 4 - \ln 3\right)$

$l = 0.406844$

Explanation:

$l = \int \sqrt{{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

Solve for $\frac{\mathrm{dx}}{\mathrm{dt}}$ given $x = \ln t$

$\frac{\mathrm{dx}}{\mathrm{dt}} = \left(\frac{1}{t}\right)$

Solve for $\frac{\mathrm{dy}}{\mathrm{dt}}$ given $y = 5 - \ln t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = - 1 \cdot \frac{1}{t}$

Solve for the length $l$

$l = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

$a = 3$ and $b = 4$

$l = {\int}_{3}^{4} \sqrt{{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

$l = {\int}_{3}^{4} \sqrt{{\left(- \frac{1}{t}\right)}^{2} + {\left(\frac{1}{t}\right)}^{2}} \mathrm{dt}$

$l = {\int}_{3}^{4} \sqrt{\frac{2}{t} ^ 2} \mathrm{dt}$

$l = {\int}_{3}^{4} \frac{\sqrt{2}}{t} \mathrm{dt}$

$l = \sqrt{2} \cdot \left(\ln 4 - \ln 3\right)$

$l = 0.406844$

God bless....I hope the explanation is useful.
user
The text contains a mathematical question about finding the arc length of a parametric function, and it provides a detailed solution with the final answer.

TYPE: SINGLE_QUESTION_WITH_ANSWER