Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (e)

MATH100 December 2011

  •  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 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q5 (d)  •  Q5 (e)  •  Q6  •  Q7  •  Q8  •  

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Question 01 (e)

Short-Answer Questions. Put your answer in the box provided but show your work also. Each question is worth 3 marks, but not all questions are of equal difficulty. Full marks will be given for correct answers placed in the box, but at most 1 mark will be given for incorrect answers. Unless otherwise stated, it is not necessary to simplify your answers in this question. Find an equation of the tangent line to the curve $y=x^{3.5}-e^{3.5}$ at the point (e, 0).

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
How do you find the slope of the tangent line to a given function at a specific point?

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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. To find the equation of the tangent line to y = x3.5 - e3.5 at the point (e, 0 ), we must first find the derivative of the curve, which is the slope of the tangent line at that point. First we use the difference rule to separate the two terms. ${\frac {d}{dx}}(x^{3.5}-e^{3.5})={\frac {d}{dx}}(x^{3.5})-{\frac {d}{dx}}(e^{3.5})$ The next step is to recognize that e3.5 is a constant, so we know that its derivative is equal to 0 and we can remove that term altogether from the equation. Then we use to power rule to differentiate x3.5 : ${\frac {d}{dx}}(x^{3.5})=3.5x^{3.5-1}=3.5x^{2.5}$ By plugging in x = e into the derivative, we find the tangent line at the desired point (e, 0), we get the slope m of the tangent line of equation y-y0 = m(x - x0) $\displaystyle m=3.5(e^{2.5})$ From this we can conclude that the equation of the line is: $\displaystyle y-y_{0}=m(x-x_{0})\quad \Rightarrow \quad y-0=3.5e^{2.5}(x-e)$ Simplifying, we get $\displaystyle y=3.5e^{2.5}(x-e)$ or $\displaystyle y=3.5e^{2.5}x-3.5e^{3.5}$

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