Find the equation of the normal to
the curve 𝑦 equals eight cos 𝑥 minus three sec 𝑥 at 𝑥 equals 𝜋 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 𝑥 is equal to 𝜋 by three. We can find the 𝑦-value at this
point by substituting 𝑥 equals 𝜋 by three into the equation of the curve. cos of 𝜋 by three is one-half, so
sec of 𝜋 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 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 𝑥 is measured in radians. Firstly, the derivative with
respect to 𝑥 of cos 𝑥 is negative sin 𝑥. And secondly, the derivative with
respect to 𝑥 of sec 𝑥 is sec 𝑥 tan 𝑥. 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 d𝑥 is equal to eight multiplied by negative sin 𝑥 minus three
multiplied by sec 𝑥 tan 𝑥. So we have the general slope
function of the curve. We need to evaluate this at the
point where 𝑥 is equal to 𝜋 by three. We have negative eight sin 𝜋 by
three minus three sec 𝜋 by three tan 𝜋 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 𝑥 equals 𝜋 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 𝑥 minus 𝜋 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 𝑥 over 30 minus root three 𝜋 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 𝑥 equals 𝜋 by three is 𝑦 minus
root three 𝑥 over 30 plus root three 𝜋 over 90 plus two equals zero.