Science:Math Exam Resources/Courses/MATH110/December 2011/Question 08 (a)

MATH110 December 2011

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Question 08 (a)
Let $y=(ax+b)^{7}$, where a and b are constant. (a) Suppose that a ≠ 0. Find the equation of the line tangent to the curve at $x={\frac {b}{a}}$.

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
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.

Checking a solution serves two purposes: helping you if, after having used the hint, 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. If a is different than zero, then the given curve is a polynomial of degree 7. Let $\displaystyle f(x)=(ax+b)^{7}$ then its derivative (using the chain rule) is $\displaystyle f'(x)=7(ax+b)^{6}\cdot a=7a(ax+b)^{6}$ and so the slope of the tangent line to this curve at the point x = b/a is $\displaystyle f'(b/a)=7a(a\left({\frac {b}{a}}\right)+b)^{6}=7a(2b)^{6}=2^{6}\cdot 7ab^{6}$ And the y coordinate of that point is $f(b/a)$ which is $\displaystyle f(b/a)=(a\left({\frac {b}{a}}\right)+b)^{7}=(2b)^{7}=2^{7}b^{7}$ So using the standard equation $\displaystyle y=m(x-x_{0})+y_{0}$ for a line of slope m and going through the point (x0, y0) we obtain $\displaystyle y=2^{6}\cdot 7ab^{6}(x-{\frac {b}{a}})+2^{7}b^{7}$ which you can simplify a bit and obtain $\displaystyle y=2^{6}\cdot 7ab^{6}x-2^{6}\cdot 5b^{7}$ or if you compute things out $\displaystyle y=448ab^{6}x-320b^{7}$ You don't need to do the last two steps during a test usually.

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