Video: Finding the Equation of the Normal to a Curve Defined Implicitly at a Given Point Involving Using the Product Rule

Find the equation of the normal to the curve π‘₯²𝑦² βˆ’ 4π‘₯ + 2𝑦 βˆ’ 20 = 0 at the point (1, 4).

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Video Transcript

Find the equation of the normal to the curve π‘₯ squared times 𝑦 squared minus four π‘₯ plus two 𝑦 minus 20 is equal to zero at the point one, four.

The question wants us to find the equation of a normal to a curve defined implicitly at the point one, four. To find the equation of a normal, we recall 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 has a slope negative one divided by π‘š. This is because the tangent and the normal lines are perpendicular to each other.

In fact, this tells us if the tangent line is horizontal, then the normal line will be vertical, and vice versa. So we can use the slope of the tangent to find the slope of our normal.

To find the slope of our tangent at the point one, four, we’re going to use differentiation. We differentiate both sides of our equation with respect to π‘₯ to help us find an expression for the slope d𝑦 by dπ‘₯. This gives us the derivative of π‘₯ squared 𝑦 squared minus four π‘₯ plus two 𝑦 minus 20 with respect to π‘₯ is equal to the derivative of zero with respect to π‘₯.

We can differentiate each term separately. We know the derivative of zero with respect to π‘₯ is just equal to zero. The derivative of negative 20 with respect to π‘₯ is also equal to zero. And the derivative of negative four π‘₯ with respect to π‘₯ is equal to negative four. To differentiate the remaining two terms, we recall that 𝑦 is a function of π‘₯. So we can differentiate these by using the chain rule.

The chain rule tells us since 𝑦 is a function of π‘₯, the derivative of 𝑓 of 𝑦 with respect to π‘₯ is equal to the derivative of 𝑓 with respect to 𝑦 multiplied by the derivative of 𝑦 with respect to π‘₯. We can use this to find the derivative of two 𝑦 with respect to π‘₯. It’s equal to the derivative of two 𝑦 with respect to 𝑦 multiplied by d𝑦 by dπ‘₯. And we know the derivative of two 𝑦 with respect to 𝑦 is just equal to two.

We see the final term we need to differentiate is the product of two functions. So we’ll need to use the product rule. The product rule tells us the derivative of the product of two functions 𝑒 and 𝑣 with respect to π‘₯ is equal to 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯. So we can use the product rule and the chain rule together to differentiate π‘₯ squared 𝑦 squared with respect to π‘₯.

Applying the product rule gives us π‘₯ squared times the derivative of 𝑦 squared with respect to π‘₯ plus 𝑦 squared times the derivative of π‘₯ squared with respect to π‘₯. We can differentiate 𝑦 squared with respect to π‘₯ by using the chain rule. It’s the derivative of 𝑦 squared with respect to 𝑦. That’s two 𝑦 multiplied by d𝑦 by dπ‘₯. And we can evaluate the derivative of π‘₯ squared with respect to π‘₯ as two π‘₯.

So we have two π‘₯ squared 𝑦 d𝑦 by dπ‘₯ plus two π‘₯ 𝑦 squared minus four plus two d𝑦 by dπ‘₯ plus zero is equal to zero. And we want to find an expression for the slope of our tangents. So we need to rewrite this expression to have d𝑦 by dπ‘₯ as the subject.

First, we’ll subtract two π‘₯𝑦 squared and add four to both sides of the equation. Then we’ll take out the shared factor of d𝑦 by dπ‘₯ from the remaining two terms. This gives us two π‘₯ squared 𝑦 plus two multiplied by d𝑦 by dπ‘₯ is equal to four minus two π‘₯𝑦 squared.

Finally, we can divide both sides of our equation by two π‘₯ squared 𝑦 plus two to get d𝑦 by dπ‘₯ is equal to four minus two π‘₯𝑦 squared all divided by two π‘₯ squared 𝑦 plus two. We want to find the equation of the normal to the curve at the point one, four. That’s when π‘₯ is equal to one and 𝑦 is equal to four. So we can substitute π‘₯ is equal to one and 𝑦 is equal to four into our expression for d𝑦 by dπ‘₯ to find the slope of the tangent to the curve at the point one, four. It’s equal to four minus two times one times four squared all divided by two times one squared times four plus two.

We can evaluate this to get an answer of negative 14 divided by five. However, this is the slope of the tangent to the curve at this point. We want the slope of the normal to the curve at this point. We can do this by finding negative one times the reciprocal of this value. The reciprocal of negative 14 divided by five is negative five divided by 14. We then multiply this by negative one to just get five divided by 14.

Since we’ve found the slope of our normal line to be five divided by 14, we can write it in the standard form for a line. 𝑦 is equal to five divided by 14π‘₯ plus 𝑏. And from the question, we know that our normal line must pass through the point one, four. Since our line passes through the point one, four, we can substitute π‘₯ is equal to one and 𝑦 is equal to four to find the value of 𝑏.

We subtract five divided by 14 from both sides. And we see that 𝑏 is equal to 51 divided by 14. Therefore, by using the fact 𝑏 is equal to 51 over 14 and then rearranging our equation. We have the equation for the normal to the curve π‘₯ squared 𝑦 squared minus four π‘₯ plus two 𝑦 minus 20 is equal to zero at the point one, four is 𝑦 minus five π‘₯ over 14 minus 51 over 14 is equal to zero.

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