Video Transcript
Find the equation of the normal to the curve 𝑦 equals negative six cot 𝑥 minus five sec squared 𝑥 plus seven at 𝑥 equals 𝜋 by four.
Remember, the normal to the curve is the line that’s perpendicular to the tangent to the curve at that point. This means if we can find the slope of the tangent to the curve at the point 𝑥 equals 𝜋 by four, then we can in turn use this to find the slope of the normal. And if two lines are perpendicular, we know that the product of their slopes, 𝑚 sub one times 𝑚 sub two, equals negative one. Once we have the slope of the normal, we can then use the general formula for the equation of a straight line. 𝑦 minus 𝑦 sub one equals 𝑚 times 𝑥 minus 𝑥 sub one, where 𝑚 is the value of the slope and 𝑥 sub one, 𝑦 sub one is the coordinate of the point on that line. So how are we going to find the slope of the tangent to the curve first?
Well, we know that to find the slope or the gradient of the tangent to the curve, we differentiate the equation of the curve and then evaluate it at that point. So, if we let 𝑚 sub one be the gradient or slope of the tangent, then we need to differentiate 𝑦 with respect to 𝑥 and evaluate that at 𝑥 equals 𝜋 by four. Now, we can differentiate term by term. First, we know that the derivative of cot 𝑥 with respect to 𝑥 is negative csc squared 𝑥. And the derivative of sec squared 𝑥 is two sec squared 𝑥 tan 𝑥.
Now, of course, we could use either the product rule or-or the chain rule to derive this directly. Finally, if 𝑎 is a real constant, then the derivative of 𝑎 with respect to 𝑥 is simply zero. This allows us to differentiate 𝑦 with respect to 𝑥 term by term. d𝑦 by d𝑥 is negative six times negative csc squared 𝑥 minus five times two sec squared 𝑥 tan 𝑥 plus zero. And that simplifies to six csc squared 𝑥 minus 10 sec squared 𝑥 tan 𝑥.
Now, to find the slope of the tangent, we just need to substitute 𝑥 equals 𝜋 by four into this equation. That gives us six csc squared 𝜋 by four minus 10 sec squared 𝜋 by four times tan 𝜋 by four. By using the identity csc equals one over sin and sec equals one over cos, we can rewrite this as six over sin squared 𝜋 by four minus 10 tan 𝜋 by four over cos squared 𝜋 by four. Both sin and cos of 𝜋 by four are root two over two, whilst tan of 𝜋 by four is one. So we get six over root two over two squared minus 10 over root two over two squared. But then if we square root two over two, we get a half. And of course dividing by one-half is the same as multiplying by two. So we get 12 minus 20, which is equal to negative eight.
We now know the slope of the tangent to the curve at the point we’re interested in. So we’ll use the fact that the product of the slope of the tangent and the slope of the normal is negative one to work out the slope of the normal to the curve. Defining the slope of the normal to be 𝑚 sub two, and we get negative eight times 𝑚 sub two equals negative one. And if we divide through by negative eight, we solve to find the slope of the normal. It’s one-eighth.
We now know the slope and we know the 𝑥-value of the coordinate of the point where this meets the curve. We still need to find the 𝑦-coordinate though. So we’re going to substitute 𝑥 equals 𝜋 by four into the original equation. When we do, we get negative six times cot of 𝜋 by four minus five sec squared 𝜋 by four plus seven. We can think about this as negative six over tan of 𝜋 by four minus five over cos squared of 𝜋 by four plus seven, which is equal to negative nine.
So we now know the slope of the normal, it’s one-eighth, and we know the 𝑥- and 𝑦-coordinates of the points where it meets the curve. We substitute all of this into our equation for a straight line. And we get 𝑦 minus negative nine equals one-eighth times 𝑥 minus 𝜋 by four. Then, distributing our parentheses, and we get 𝑦 plus nine equals 𝑥 over eight minus 𝜋 by 32. And then we’ll set this equal to zero. We subtract 𝑥 over eight from both sides and then add 𝜋 by 32. The equation of the normal to the curve is negative 𝑥 over eight plus 𝑦 plus nine plus 𝜋 by 32 equals zero.
