MATH110 December 2011
• Q1 (a) • Q1 (b) • Q1 (c) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q5 (c) • Q6 • Q7 (a) • Q7 (b) • Q7 (c) • Q7 (d) • Q7 (e) • Q8 (a) • Q8 (b) • Q9 • Q10 (a) • Q10 (b) • Q11 •
Question 08 (a)
Let , where a and b are constant.
(a) Suppose that a ≠ 0. Find the equation of the line tangent to the curve at
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!
If it looks like keeping track of a lot of symbols and having an answer that will depend on both a and b, then you are on the right track.
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If a is different than zero, then the given curve is a polynomial of degree 7. Let
then its derivative (using the chain rule) is
and so the slope of the tangent line to this curve at the point x = b/a is
And the y coordinate of that point is which is
So using the standard equation
for a line of slope m and going through the point (x0, y0) we obtain
which you can simplify a bit and obtain
or if you compute things out
You don't need to do the last two steps during a test usually.
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