Video: Finding the Equation of the Normal to the Curve of a Function Defined by Parametric Equations Containing Reciprocal Trigonometric Functions

Find the equation of the normal to the curve π‘₯ = 3 csc πœƒ, 𝑦 = 2 cot πœƒ + 6 csc πœƒ at πœƒ =(4πœ‹)/3.

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Video Transcript

Find the equation of the normal to the curve π‘₯ is equal to three times the csc to πœƒ and 𝑦 is equal to two times the cot to πœƒ plus six times the csc to πœƒ at πœƒ is equal to four πœ‹ by three.

The question gives us a pair of parametric equations. And it wants us to find the equation of the normal to the curve at the point where πœƒ is equal to four πœ‹ by three. We recall for a pair of parametric equations π‘₯ is equal to some function 𝑓 of πœƒ and 𝑦 is equal to some function 𝑔 of πœƒ. We can find the slope of the curve d𝑦 by dπ‘₯ by dividing the derivative of 𝑦 with respect to πœƒ by the derivative of π‘₯ with respect to πœƒ. However, this will only give us the equation of the tangent to the curve at this point.

However, we know if the tangent to the curve at the point 𝑝 has a slope of π‘š, then the normal to the curve at the point 𝑝 will have a slope of negative one divided by π‘š. This is because our normal line will be perpendicular to the tangent line. In fact, this means if our tangent line was horizontal, then our normal line must be vertical. And this’s true vice versa. So, to find the slope of our normal, we’ll start by finding the derivative of 𝑦 with respect to πœƒ. That’s the derivative of two cot of πœƒ plus six csc of πœƒ with respect to πœƒ.

To evaluate this derivative, we recall two standard derivatives of trigonometric functions. The derivative of the cot of πœƒ with respect to πœƒ is equal to negative the csc squared of πœƒ. And the derivative of the csc to πœƒ with respect to πœƒ is equal to negative the cot of πœƒ multiplied by the csc of πœƒ. Applying this, we have d𝑦 by dπœƒ is equal to negative two times the csc squared of πœƒ minus six times the cot of πœƒ times the csc of πœƒ.

We can do the same to find dπ‘₯ by dπœƒ. It’s the derivative of three times the csc of πœƒ with respect to πœƒ. And differentiating this gives us negative three times the cot of πœƒ times the csc of πœƒ. We want to find the slope of our tangent when πœƒ is equal to four πœ‹ by three. Instead of dividing d𝑦 by dπœƒ by dπ‘₯ by dπœƒ and then evaluating at πœƒ is equal to four πœ‹ by three, we can evaluate both expressions at four πœ‹ by three and then divide.

We have d𝑦 by dπœƒ when πœƒ is equal to four πœ‹ by three is equal to negative two times the csc squared at four πœ‹ by three minus six times the cot of four πœ‹ by three times the csc of four πœ‹ by three. We could evaluate this by calculator. However, we should know that the sin of four πœ‹ by three is equal to negative root three over two. And the tan of four πœ‹ by three is equal to the square root of three. This is because the sin of four πœ‹ by three is just negative the sin of πœ‹ by three, which is a standard angle that we should memorize. And the tangent of four πœ‹ by three is just equal to the tangent of πœ‹ by three.

We can then take the reciprocal of both sides of both of these equations. Then, using standard trigonometric identities, we can see the csc of four πœ‹ by three is equal to negative two divided by root three. And the cot of four πœ‹ by three is equal to one divided by the square root of three. We can then evaluate our expression for d𝑦 by dπœƒ to get negative two times negative two over root three squared minus six times one over root three times negative two over root three.

Evaluating this expression, we get four divided by three. We can do the same to get dπ‘₯ by dπœƒ when πœƒ is equal to four πœ‹ by three. We get negative three multiplied by one over square root of three multiplied by negative two divided by root three. Which, if we evaluate, we just get the answer of two.

Remember that d𝑦 by dπ‘₯ is equal to d𝑦 by dπœƒ divided by dπ‘₯ by dπœƒ. So, if we divide four-thirds by two, we will get the slope of our curve when πœƒ is equal to four πœ‹ by three. So, we have the slope d𝑦 by dπ‘₯ when πœƒ is equal to four πœ‹ by three is equal to four-thirds divided by two, which is just equal to two-thirds. And we know this is the slope of the tangent to the curve. So, to find the slope of the normal, we need to take negative the reciprocal of this value.

So, the slope of our normal is equal to negative the reciprocal of two-thirds, which is just equal to negative three divided by two. We recall we can find the equation of a line which passes through the point π‘₯ one, 𝑦 one with a slope of π‘š by using 𝑦 minus 𝑦 one is equal to π‘š times π‘₯ minus π‘₯ one. Where π‘š is the slope of our line. We’ve already found the slope of our normal line. It has a slope π‘š of negative three divided by two. To find the point on our line, we need to substitute πœƒ is equal to four πœ‹ by three into our parametric equations.

Substituting πœƒ is equal to four πœ‹ by three gives us that π‘₯ is equal to three times the csc of four πœ‹ by three. Which is equal to three times negative two divided by root three, which we can simplify to be negative six divided by root three. We can do the same for 𝑦. We get two times the cot of four πœ‹ by three plus six times the csc of four πœ‹ by three. Which is two times one over root three plus six times negative two divided by root three.

Simplifying and combining these fractions, we get negative 10 divided by the square root of three. We set π‘₯ one equal to negative six divided by the square root of three and 𝑦 one to be negative 10 divided by the square root of three. Substituting these values into our equation for a straight line gives us 𝑦 minus negative 10 over root three is equal to negative three over two times π‘₯ minus negative six over root three.

Simplifying this expression and then multiplying out the parentheses, we get 𝑦 plus 10 over root three is equal to negative three π‘₯ over two minus nine divided by root three. We can then write all of our terms on the same side of the equation. This gives us 𝑦 plus three π‘₯ over two plus 19 over the square root of three is equal to zero. We could leave our answer like this. However, we can simplify 19 divided by root three by multiplying the numerator and the denominator by root three.

This gives us 19 root three divided by three. Then, by reordering our first two terms, we have three π‘₯ over two plus 𝑦 plus 19 root three over three is equal to zero. Therefore, we’ve shown that an equation for the normal to the curve π‘₯ is equal to three csc of πœƒ, 𝑦 is equal to two cot of πœƒ plus six csc of πœƒ at πœƒ is equal to four πœ‹ by three is three π‘₯ over two plus 𝑦 plus 19 root three over three is equal to zero.

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