Video Transcript
Determine the equation of the normal to the curve 𝑥 is equal to negative four cot of 𝜃 plus three, 𝑦 is equal to three sin squared of 𝜃 plus root two sec of 𝜃 at 𝜃 is equal to 𝜋 by four.
In this question, we’re given a pair of parametric equations, where the parametric equations involve trigonometric functions. We need to determine the equation of the normal to these parametric equations when 𝜃 is equal to 𝜋 by four. To do this, let’s start by recalling what we mean by the normal to the curve. It’s the line perpendicular to the tangent of the curve at this point, and it also passes through the point. Therefore, since we’re asked to determine the equation of the normal to this curve at 𝜃 is equal to 𝜋 by four, we can substitute 𝜃 is equal to 𝜋 by four into both of our parametric equations. This will give us the coordinates of a point which lies on our normal line.
Let’s start by determining the 𝑥-coordinate of this point. We have 𝑥 is equal to negative four cot of 𝜋 by four plus three. And we can evaluate this expression directly. First, the cot of 𝜋 by four is one over the tan of 𝜋 by four, and we know the tan of 𝜋 by four is one. So the cot of 𝜋 by four is also one. This gives us that 𝑥 is equal to negative four plus three, which is negative one.
We can now do the same to find the 𝑦-coordinate of this point. We have 𝑦 is equal to three times the sin squared of 𝜋 by four plus root two times the sec of 𝜋 by four. And we can evaluate this term by term. First, the sin of 𝜋 by four is one divided by root two. So the first term is three times one over root two squared. Next, the sec of 𝜋 by four is one divided by the cos of 𝜋 by four. And we also know the cos of 𝜋 by four is one over root two. So this term is root two times one divided by one over root two. And we can simplify this slightly. One divided by one over root two is just root two. So this gives us that 𝑦 is equal to three times one over root two squared plus root two multiplied by root two.
Now, all we need to do is simplify and evaluate this expression. First, one over root two squared is one-half, and root two multiplied by root two is two, so 𝑦 is three times one-half plus two. And we can then just evaluate this expression. We get seven over two. So we’ve now shown that our normal line passes through the point with coordinates negative one, seven over two.
And we can then recall to find the equation of a line, we need to find a point it passes through and its slope. So all we need to do now is determine the slope of this line. And we can do this by first recalling the slope of the tangent line to this curve will be d𝑦 by d𝑥 evaluated at 𝜃 is equal to 𝜋 by four. Then, provided the slope of the tangent line is not zero, we can recall the slope of the normal line will be negative the reciprocal of this value. It’s negative one divided by d𝑦 by d𝑥 evaluated at 𝜃 is equal to 𝜋 by four.
So to determine the slope of the normal line at this point, we’re going to need to find an expression for d𝑦 by d𝑥. And since we’re given a parametric curve, we’re going to need to do this by using parametric differentiation, which is just an application of the chain rule. If 𝑥 is a differentiable function in 𝜃 and 𝑦 is a differentiable function in 𝜃, then we can rearrange the chain rule to show d𝑦 by d𝑥 is equal to d𝑦 by d𝜃 divided by d𝑥 by d𝜃, provided d𝑥 by d𝜃 is not equal to zero at the point. So we’re going to need to differentiate both of our parametric equations. Let’s start with the parametric equation in 𝑥.
Differentiating both sides of this equation with respect to 𝜃, we get d𝑥 by d𝜃 is equal to the derivative of negative four cot of 𝜃 plus three with respect to 𝜃. And we can evaluate this derivative term by term by recalling the derivative of the cot of 𝜃 with respect to 𝜃 is negative csc squared of 𝜃. And therefore, since the derivative of the constant three is zero, d𝑥 by d𝜃 is four csc squared of 𝜃.
We now want to do the same to find the derivative of 𝑦 with respect to 𝜃. We differentiate both sides of this equation with respect to 𝜃. This gives us d𝑦 by d𝜃 is equal to the derivative of three sin squared of 𝜃 plus root two sec of 𝜃 with respect to 𝜃. We want to differentiate this term by term. Let’s start with the first term, how to differentiate three sin squared of 𝜃 with respect to 𝜃. Since this term is the composition of two functions, a square and a sine function, we’ll do this by using the chain rule. First, we multiply by our constant. Next, we multiply by the derivative of the inner function. That’s the derivative of the sin of 𝜃 with respect to 𝜃. And finally, we need to differentiate the outer function and evaluate this at the inner function. This gives us two sin of 𝜃.
We can then simplify this by noting the derivative of the sin of 𝜃 with respect to 𝜃 is the cos of 𝜃. Substituting this into our expression and then noting three times two is six, we get six times the sin of 𝜃 multiplied by the cos of 𝜃. To differentiate the second term, we just need to note the derivative of the sec of 𝜃 with respect to 𝜃 is the sec of 𝜃 multiplied by the tan of 𝜃. So we can differentiate this term to get root two sec of 𝜃 times the tan of 𝜃. And hence, d𝑦 by d𝜃 is six sin of 𝜃 cos of 𝜃 plus root two sec of 𝜃 tan of 𝜃.
We’ve now found expressions for d𝑦 by d𝜃 and d𝑥 by d𝜃. We can use this to determine an expression for d𝑦 by d𝑥. We clear some space and then substitute these expressions into our formula for d𝑦 by d𝑥. This gives us that d𝑦 by d𝑥 is equal to six sin 𝜃 times cos of 𝜃 plus root two sec of 𝜃 multiplied by the tan of 𝜃 all divided by four csc squared of 𝜃. We can then use this to determine the slope of the tangent line when 𝜃 is equal to 𝜋 by four. We just need to substitute 𝜃 is equal to 𝜋 by four into this equation. Doing this and clearing some space, we get d𝑦 by d𝑥 at 𝜃 is equal to 𝜋 by four is equal to six times the sin of 𝜋 by four multiplied by the cos of 𝜋 by four plus root two times the sec of 𝜋 by four times the tan of 𝜋 by four all divided by four times the csc squared of 𝜋 by four.
We can now start evaluating this expression. First, in our numerator, both the sin of 𝜋 by four and the cos of 𝜋 by four are equal to one over root two. Next, also in the numerator, the sec of 𝜋 by four is one over the cos of 𝜋 by four, which is equal to root two. And the tan of 𝜋 by four is just equal to one. Finally, in the denominator, the csc of 𝜋 by four is equal to one over the sin of 𝜋 by four, which is root two. Therefore, d𝑦 by d𝑥 at 𝜃 is 𝜋 by four, which is the slope of the tangent line to the curve when 𝜃 is 𝜋 by four, is given by six times one over root two multiplied by one over root two plus root two times root two times one all divided by four times root two squared.
Now, all that’s left to do is evaluate this expression. One over root two times one over root two is a half. Multiply this by six, and we get three. And root two times root two is two, and four times root two squared is eight. So we get three plus two all over eight, which is five over eight. And this is the slope of the tangent line at this point. But remember, we want the slope of the normal line at this point. So we need to take negative the reciprocal of this value, and this is negative eight over five. So our normal line has a slope of negative eight over five and it passes through the point with coordinates negative one, seven over two, which means we can find an equation for the normal line by substituting these values into the point–slope form of the equation of a line.
Remember, that’s the form 𝑦 minus 𝑦 sub one is equal to 𝑚 times 𝑥 minus 𝑥 sub one, where the line has slope 𝑚 and passes through the point with coordinates 𝑥 sub one, 𝑦 sub one. And our normal line has a slope of negative eight over five and passes through the point with coordinates negative one, seven over two. So we substitute these values into the equation. We get 𝑦 minus seven over two is equal to negative eight over five multiplied by 𝑥 minus negative one.
Now, all that’s left to do is simplify this equation. First, 𝑥 minus negative one is equal to 𝑥 plus one. Next, we’re going to multiply both sides of the equation through by negative five over eight to make the coefficient of 𝑥 one. This gives us that negative five 𝑦 over eight plus 35 over 16 is equal to 𝑥 plus one. Finally, we’re going to rewrite the equation by writing all of the terms on the same side of the equation. And this gives us our final answer. The equation of the normal to the curve 𝑥 is equal to negative four cot of 𝜃 plus three and 𝑦 is equal to three sin squared of 𝜃 plus root two sec of 𝜃 at 𝜃 is equal to 𝜋 by four is 𝑥 plus five 𝑦 over eight minus 19 over 16 is equal to zero.
