Question Video: Finding the Equation of the Normal to the Curve of a Function Involving Trigonometric Functions at a Given π‘₯-Coordinate Mathematics

Find the equation of the normal to the curve 𝑦 = 8 cos π‘₯ βˆ’ 3 sec π‘₯ at π‘₯ = πœ‹/3.

03:30

Video Transcript

Find the equation of the normal to the curve 𝑦 equals eight cos π‘₯ minus three sec π‘₯ at π‘₯ equals πœ‹ by three.

To find the equation of the normal to any curve, we need to know the coordinates of a point that lies on the normal and its slope. We’re asked for the normal at the point where π‘₯ is equal to πœ‹ by three. We can find the 𝑦-value at this point by substituting π‘₯ equals πœ‹ by three into the equation of the curve. cos of πœ‹ by three is one-half, so sec of πœ‹ by three is two. We have eight multiplied by a half minus three multiplied by two. That’s four minus six, which is equal to negative two. So, the coordinates of the point at which we’re finding the normal are πœ‹ by three, negative two.

Next, we need to find the slope of the normal. Recall that the normal at any point on a curve is perpendicular or orthogonal to the tangent at that same point. And hence, their slopes are the negative reciprocals of one another. The slope of the tangent is the same as the slope of the curve itself. And we can find the slope function of the curve using differentiation. We need to recall two general rules for differentiating trigonometric functions. These rules only apply when the angle π‘₯ is measured in radians. Firstly, the derivative with respect to π‘₯ of cos π‘₯ is negative sin π‘₯. And secondly, the derivative with respect to π‘₯ of sec π‘₯ is sec π‘₯ tan π‘₯. We should be familiar with the derivatives of the three trigonometric functions sin, cos, and tan as well as their reciprocals csc, sec, and cot.

Applying these results then, we have that d𝑦 by dπ‘₯ is equal to eight multiplied by negative sin π‘₯ minus three multiplied by sec π‘₯ tan π‘₯. So we have the general slope function of the curve. We need to evaluate this at the point where π‘₯ is equal to πœ‹ by three. We have negative eight sin πœ‹ by three minus three sec πœ‹ by three tan πœ‹ by three. Evaluating on a calculator or recalling these results from memory, we have negative eight multiplied by root three over two minus three multiplied by two multiplied by root three. That’s negative four root three minus six root three, which simplifies to negative 10 root three.

Remember though that this is the slope of the tangent at the point where π‘₯ equals πœ‹ by three. The slope of the normal is the negative reciprocal of this. We can cancel the negatives and multiply both the numerator and denominator by root three in order to rationalize the denominator. So we find that the slope of the normal is root three over 30.

Finally, we’re able to calculate the equation of the normal by substituting the coordinates of the point and the slope we found into the point–slope form of the equation of a straight line. We have 𝑦 minus negative two equals root three over 30 multiplied by π‘₯ minus πœ‹ by three. Simplifying on the left-hand side and then distributing the parentheses on the right-hand side, we have 𝑦 plus two is equal to root three π‘₯ over 30 minus root three πœ‹ over 90. Finally, we’ll group all the terms on the same side of the equation. And we have that the equation of the normal to the given curve at the point where π‘₯ equals πœ‹ by three is 𝑦 minus root three π‘₯ over 30 plus root three πœ‹ over 90 plus two equals zero.

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