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

MATH110 December 2010

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[hide] Question 05 (c)
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. Consider the curve $y=\sin x$ . Find constants b and c such that the curve $y=x^{2}+bx+c$ intersects it orthogonally at $x=0$ .

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[show] Hint
To solve this question, you will need to meet the two conditions given in the definition of orthogonal intersection. It might be easiest to start with the second condition - that the tangent slopes of each curve be perpendicular to each other. What is the tangent slope of the first function you're given? What does the tangent slope of the second function have to be, if it's going to be the negative reciprocal of that first slope?

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[show] Solution
To find a solution for $b,c$ , we need to satisfy two conditions: (1) $\sin x$ and $x^{2}+bx+c$ must intersect at $x=0$ and (2) the tangent line of $\sin x$ at $x=0$ must be perpendicular to the tangent line of $x^{2}+bx+c$ at $x=0$ . For this problem it doesn't matter which condition you test first; in this case, we will start with the second condition and then proceed to the first condition. We begin by finding the slope of the tangent line to each curve at $x=0$ . For $\sin x$ the derivative is $\cos x$ and so the slope at $x=0$ is $\cos(0)=1$ . For $x^{2}+bx+c$ the derivative is $2x+b$ and so the slope at $x=0$ is $2(0)+b=b$ . To satisfy condition (2), these two slopes must be perpendicular, i.e. negative reciprocals of each other. Since the tangent slope of $\sin x$ is 1, this means we need the tangent slope of $x^{2}+bx+c$ to equal to -1. However, we know the tangent slope of $x^{2}+bx+c$ is $b$ , so this means $b=-1$ . Having found $b$ we now turn to the first condition of the definition to find $c$ . The first condition requires that $\sin x$ and $x^{2}+bx+c$ intersect at $x=0$ . This means that $\sin(0)$ must equal $(0)^{2}+(-1)(0)+c$ . In equations, this is: ${\begin{aligned}\sin(0)&=(0)^{2}+-1(0)+c\\0&=c\end{aligned}}$ So $c=0$ . Thus the curve $y=x^{2}-x$ satisfies both of our conditions and intersects orthogonally the curve $y=\sin x$ at $x=0$ .