text: 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.
explanation: The document contains a math question and a detailed answer, including the step-by-step solution and the final result.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: 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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a detailed answer, including the step-by-step solution and the final result.
  
  Label: question-and-answer