Science:Math Exam Resources/Courses/MATH110/December 2010/Question 05 (b)

MATH110 December 2010

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Question 05 (b)
Two curves are said to intersect orthogonally at x =a if (i) they intersect at x = a and (ii) their tangent lines are perpendicular at x = a. Are the tangent lines of the curves in part (a) perpendicular at $x=3$? Justify your answer.

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
Tangent lines are perpendicular if their slopes are perpendicular. How can you find the slopes of tangent lines at a point? If the slopes are perpendicular, what will the relationship between them be?

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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. First we need to calculate the slope of the tangent line to each curve at $x=3$. We do this by taking the derivative and then plugging in $x=3$. For the first curve we use the chain rule to find the derivative: ${\begin{aligned}{\frac {dy}{dx}}&={\frac {1}{6}}\cdot {\frac {1}{2}}(2x^{2}+7)^{-{\frac {1}{2}}}\cdot 4x\\&={\frac {4x}{12{\sqrt {2x^{2}+7}}}}\\&={\frac {x}{3{\sqrt {2x^{2}+7}}}}\end{aligned}}$ Plugging in $x=3$ to find the slope we get: ${\frac {3}{3{\sqrt {2(3)^{2}+7}}}}={\frac {1}{\sqrt {25}}}={\frac {1}{5}}$. For the second curve we simply use the power rule to calculate the derivative: ${\frac {dy}{dx}}=-4x+7$ Plugging in $x=3$ gives a slope of: $\displaystyle -4(3)+7=-5$. We know that two slopes are perpendicular if they are negative reciprocals of each other, i.e. if one slope is ${\frac {a}{b}}$ then the other slope must be $-{\frac {b}{a}}$. For this problem, this relationship is true: the first slope, ${\frac {1}{5}}$ is the negative reciprocal of $-5$. Thus the tangent lines to these two curves are perpendicular at $x=3$.

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