Question Video: Finding the Equation of the Normal to a Curve Defined Implicitly at a Given Point Mathematics

Find the equation of the normal to the curve 3𝑦² βˆ’ 9𝑦π‘₯ + 7π‘₯Β² = 1 at the point (βˆ’1, βˆ’1).

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Video Transcript

Find the equation of the normal to the curve three 𝑦 squared minus nine 𝑦π‘₯ plus seven π‘₯ squared equals one at the point negative one, negative one.

The curve we’ve been given has been defined implicitly. It is a function of both π‘₯ and 𝑦 and cannot be easily rearranged to a form in which we have 𝑦 as a function of π‘₯. To find the equation of the normal to this curve, we’re going to use the point–slope form of the equation of a straight line. That’s 𝑦 minus 𝑦 one equals π‘š π‘₯ minus π‘₯ one. We know the coordinates of a point on the normal, the point negative one, negative one. But we need to calculate its slope. We recall that a normal to a curve is perpendicular to the tangent to the curve at that point. So first, we must calculate the slope of the tangent. To do this, we’re going to need to find the slope function of the curve. So we need to recall how we differentiate a function which is defined implicitly.

Implicit differentiation is an application of the chain rule. Recall that if 𝑦 is a function of 𝑒 and 𝑒 is a function of π‘₯, then the chain rule states that d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑒 multiplied by d𝑒 by dπ‘₯. So the derivative of 𝑦 with respect to π‘₯ is found by taking the derivative of 𝑦 with respect to 𝑒 and then multiplying this by the derivative of 𝑒 with respect to π‘₯. Let’s consider then how we can apply this to the equation of our curve. We’ll do this term by term, starting with the first term three 𝑦 squared. As three 𝑦 squared is a function of 𝑦 and we can consider 𝑦 to be a function of π‘₯, by the chain rule, we have that the derivative with respect to π‘₯ of three 𝑦 squared is equal to the derivative with respect to 𝑦 of three 𝑦 squared multiplied by d𝑦 by dπ‘₯.

By the power rule of differentiation, the derivative with respect to 𝑦 of three 𝑦 squared is simply six 𝑦. So we find that the derivative with respect to π‘₯ of three 𝑦 squared is six 𝑦 d𝑦 by dπ‘₯. So, we’ve differentiated the first term in our equation implicitly. Differentiating the third term is straightforward because it is a function of π‘₯ only. By the power rule of differentiation, the derivative with respect to π‘₯ of seven π‘₯ squared is 14π‘₯. And differentiating the term on the right-hand side of the equation is straightforward because it is a constant. And we know that the derivative of any constant with respect to π‘₯ is simply zero.

We have one term left to differentiate, negative nine 𝑦π‘₯. And this is a little more complicated as it is a product involving both 𝑦 and π‘₯. We’ll need to use implicit differentiation. But we also need to recall the product rule. This states that for two differentiable functions 𝑒 and 𝑣, the derivative with respect to π‘₯ of their product 𝑒𝑣 is equal to 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯. We’ll therefore let 𝑒 equal negative nine 𝑦 and 𝑣 equal π‘₯. We then need to find their individual derivatives with respect to π‘₯. d𝑣 by dπ‘₯ is straightforward because 𝑣 is a function of π‘₯ only. If 𝑣 is equal to π‘₯, then d𝑣 by dπ‘₯ is equal to one.

For d𝑒 by dπ‘₯ though, as 𝑒 is a function of 𝑦, we’re going to need to use implicit differentiation again. By the chain rule, we have that the derivative with respect to π‘₯ of negative nine 𝑦 is equal to the derivative with respect to 𝑦 of negative nine 𝑦 multiplied by d𝑦 by dπ‘₯. The derivative with respect to 𝑦 of negative nine 𝑦 is simply negative nine. So we have that d𝑒 by dπ‘₯ is equal to negative nine d𝑦 by dπ‘₯. We then need to substitute each of these expressions into the product rule. We have that the derivative with respect to π‘₯ of negative nine 𝑦π‘₯ is equal to 𝑒 times d𝑣 by dπ‘₯. That’s negative nine 𝑦 multiplied by one. Then we add 𝑣 times d𝑒 by dπ‘₯. That’s π‘₯ multiplied by negative nine d𝑦 by dπ‘₯. And this will simplify to negative nine 𝑦 minus nine π‘₯ d𝑦 by dπ‘₯.

So we’ve now differentiated the entire equation, giving six 𝑦 d𝑦 by dπ‘₯ minus nine 𝑦 minus nine π‘₯ d𝑦 by dπ‘₯ plus 14π‘₯ is equal to zero. We want to evaluate the slope function at a particular point. So first, we need to rearrange this equation to give an explicit expression for d𝑦 by dπ‘₯. In other words, we need to make d𝑦 by dπ‘₯ the subject of this equation. We begin by collecting the terms that do involve d𝑦 by dπ‘₯ on one side of the equation and the terms that don’t on the other. So we have six 𝑦 d𝑦 by dπ‘₯ minus nine π‘₯ d𝑦 by dπ‘₯ is equal to nine 𝑦 minus 14π‘₯. We can then factor the left-hand side of the equation to give six 𝑦 minus nine π‘₯ multiplied by d𝑦 by dπ‘₯ is equal to nine 𝑦 minus 14π‘₯.

And to find an expression for d𝑦 by dπ‘₯ which is in terms of both π‘₯ and 𝑦, we divide both sides of the equation by six 𝑦 minus nine π‘₯. So we have d𝑦 by dπ‘₯ is equal to nine 𝑦 minus 14π‘₯ over six 𝑦 minus nine π‘₯. We have an expression for the general slope function of the curve. But we need to evaluate the slope at a particular point, the point negative one, negative one. We therefore need to substitute the value negative one for both π‘₯ and 𝑦. We have negative nine plus 14 in the numerator and negative six plus nine in the denominator, which simplifies to the fraction five over three.

Remember though that this is the slope of the tangent to the curve at the point negative one, negative one, whereas we’re looking to find the equation of the normal. But we know that the slope of the normal is the negative reciprocal of the slope of the tangent. So we have that the slope of the normal is equal to negative three-fifths. We can invert the fraction and change the sign.

As we now know the coordinates of a point on the normal and its slope, we can find its equation using the point–slope form of the equation of a straight line. We have 𝑦 minus negative one is equal to negative three-fifths π‘₯ minus negative one. Of course, that’s just 𝑦 plus one equals negative three-fifths π‘₯ plus one. We can multiply through by five and distribute the parentheses on the right-hand side to give five 𝑦 plus five equals negative three π‘₯ minus three. And finally, we’ll group all of the terms on the left-hand side of the equation. Using implicit differentiation then, we found that the equation of the normal to the given curve at the point negative one, negative one is five 𝑦 plus three π‘₯ plus eight equals zero.

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