Science:Math Exam Resources/Courses/MATH104/December 2010/Question 02 (b)

MATH104 December 2010

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q1 (k)  •  Q1 (l)  •  Q1 (m)  •  Q1 (n)  •  Q2 (a)  •  Q2 (b)  •  Q3  •  Q4  •  Q5  •  Q6  •  

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Question 02 (b)
Use a suitable linear approximation (tangent line approximation) to estimate the x-coordinate of the point on the curve whose y-coordinate is 31/16. A calculator-ready answer is enough.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The graph of the linear approximation near a point on a curve is simply the tangent line to the curve at that point.

Hint 2
You've already found the equation of the tangent line at the point (1,2) in part (a). Is the point on the curve with y-coordinate 31/16 close to (1,2)? If so, you can use the tangent line equation from part (a).

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First note that the difference between the y-coordinates of this point and (1,2) is $31/16-2=-1/16$, which is fairly small. It is probably small enough to use the tangent line from part (a) to find a reasonable approximation for the x-coordinate of this point. (We are not asked to estimate the error in this approximation.) The x-coordinate of the point on the tangent line with y-coordinate 31/16 can be found by substituting y=31/16 into the equation for the tangent line and solving for x. ${\frac {31}{16}}-2=-{\frac {7}{16}}(x-1)$ Solving for $x$, $x=1-{\frac {16}{7}}\left({\frac {31}{16}}-2\right)=1-{\frac {16}{7}}\left(-{\frac {1}{16}}\right)=1+{\frac {1}{7}}={\frac {8}{7}}$. Making a linear approximation, we conclude that the point on the curve with y-coordinate 31/16 is approximately the same as the point on the tangent line with y-coordinate 31/16. Hence, the x-coordinate of the point on the curve with y-coordinate 31/16 is approximately 8/7.

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