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

MATH100 December 2011

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[hide] 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).

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

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[show] Solution
To find the equation of the tangent line to y = x^3.5 - e^3.5 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 e^3.5 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 x^3.5 : ${\frac {d}{dx}}(x^{3.5})=3.5x^{3.5-1}=3.5x^{2.5}$ 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 $\displaystyle y=3.5e^{2.5}x-3.5e^{3.5}$