text: # What is the arclength of (t-3,t+4) on t in [2,4]?
  
  Apr 2, 2018
  
  $A = 2 \sqrt{2}$
  
  #### Explanation:
  
  The formula for parametric arc length is:
  $A = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \setminus \mathrm{dt}$
  
  We begin by finding the two derivatives:
  
  $\frac{\mathrm{dx}}{\mathrm{dt}} = 1$ and $\frac{\mathrm{dy}}{\mathrm{dt}} = 1$
  
  This gives that the arc length is:
  $A = {\int}_{2}^{4} \sqrt{{1}^{2} + {1}^{2}} \setminus \mathrm{dt} = {\int}_{2}^{4} \sqrt{2} \setminus \mathrm{dt} = {\left[\sqrt{2} t\right]}_{2}^{4} = 4 \sqrt{2} - 2 \sqrt{2} = 2 \sqrt{2}$
  
  In fact, since the parametric function is so simple (it is a straight line), we don't even need the integral formula. If we plot the function in a graph, we can just use the regular distance formula:
  
  $A = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}} = \sqrt{4 + 4} = \sqrt{8} = \sqrt{4 \cdot 2} = 2 \sqrt{2}$
  
  This gives us the same result as the integral, showing that either method works, although in this case, I'd recommend the graphical method because it is simpler.
explanation: The document contains a math question and its corresponding answer. The question asks for the arclength of a parametric curve defined by \((t-3, t+4)\) on the interval \([2, 4]\). The answer is provided as \(A = 2 \sqrt{2}\), along with a detailed explanation of the steps to derive this result using both the integral formula for arc length and the distance formula.
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 arclength of (t-3,t+4) on t in [2,4]?
  
  Apr 2, 2018
  
  $A = 2 \sqrt{2}$
  
  #### Explanation:
  
  The formula for parametric arc length is:
  $A = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \setminus \mathrm{dt}$
  
  We begin by finding the two derivatives:
  
  $\frac{\mathrm{dx}}{\mathrm{dt}} = 1$ and $\frac{\mathrm{dy}}{\mathrm{dt}} = 1$
  
  This gives that the arc length is:
  $A = {\int}_{2}^{4} \sqrt{{1}^{2} + {1}^{2}} \setminus \mathrm{dt} = {\int}_{2}^{4} \sqrt{2} \setminus \mathrm{dt} = {\left[\sqrt{2} t\right]}_{2}^{4} = 4 \sqrt{2} - 2 \sqrt{2} = 2 \sqrt{2}$
  
  In fact, since the parametric function is so simple (it is a straight line), we don't even need the integral formula. If we plot the function in a graph, we can just use the regular distance formula:
  
  $A = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}} = \sqrt{4 + 4} = \sqrt{8} = \sqrt{4 \cdot 2} = 2 \sqrt{2}$
  
  This gives us the same result as the integral, showing that either method works, although in this case, I'd recommend the graphical method because it is simpler.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and its corresponding answer. The question asks for the arclength of a parametric curve defined by \((t-3, t+4)\) on the interval \([2, 4]\). The answer is provided as \(A = 2 \sqrt{2}\), along with a detailed explanation of the steps to derive this result using both the integral formula for arc length and the distance formula.
  
  Label: question-and-answer