Question Video: Finding the Equation of the Tangent to a Curve Defined Implicitly at a Given Point Using Implicit Differentiation

Determine the equation of the tangent to the curve 6π‘₯Β³ + π‘₯𝑦² βˆ’ 5𝑦² = 0 at the point (2, 4).

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Video Transcript

Determine the equation of the tangent to the curve six π‘₯ cubed plus π‘₯𝑦 squared minus five 𝑦 squared equals zero at the point two, four.

In this question, we’ve been given a curve for which we’re trying to find the tangent at a point. We know that the tangent is a straight line. So its general equation is 𝑦 minus 𝑦 one equals π‘š times π‘₯ minus π‘₯ one. Here, π‘š is the slope of the tangent to the curve at that point. And π‘₯ one, 𝑦 one is the point it passes through. So in this case, we can say π‘₯ one is equal to two and 𝑦 one is equal to four.

But how do we find the slope of the tangent to the curve at that point? Well, usually, we’d look to differentiate our equation with respect to π‘₯. Now, we can do this term by term. But we’re going to need to be a little bit careful. We’ve been given an implicit equation, that is, an equation given in terms of both π‘₯ and 𝑦. Now, to differentiate implicitly, we use a special version of the chain rule. This says that the derivative of a function in 𝑦 with respect to π‘₯ is equal to the derivative of that function with respect to 𝑦 times d𝑦 by dπ‘₯. So let’s differentiate each term in our equation in turn.

We’ll begin by differentiating six π‘₯ cubed with respect to π‘₯. To differentiate a power term of this form, we multiply the entire term by the exponent and then reduce that exponent by one. So we get three times six π‘₯ squared. That’s 18π‘₯ squared. We’ll differentiate the third term negative five 𝑦 squared next. And we’ll deal with π‘₯𝑦 squared in a moment. To do so, we’ll use that special version of the chain rule.

We’re going to differentiate negative five 𝑦 squared with respect to 𝑦 and then multiply that by d𝑦 by dπ‘₯. Once again, we multiply the entire term by the exponent and reduce that exponent by one. So we get two times negative five 𝑦. That’s negative 10𝑦. So the derivative of negative five 𝑦 squared with respect to π‘₯ is negative 10𝑦 d𝑦 by dπ‘₯.

Now, we go back to the middle term. We need to differentiate π‘₯𝑦 squared with respect to π‘₯. And to do so, we need to notice that this itself is the product of two differentiable functions. Now, the product rule says that the derivative of the product of two differentiable functions, 𝑒 and 𝑣, is 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯. So let’s define 𝑒 to be equal to π‘₯ and 𝑣 to be equal to 𝑦 squared.

We know that d𝑒 by dπ‘₯ is simply equal to one. And once again, we use the special version of the chain rule to differentiate 𝑦 squared with respect to π‘₯. It’s the derivative of 𝑦 squared with respect to 𝑦 times d𝑦 by dπ‘₯, which is two 𝑦 d𝑦 by dπ‘₯. We substitute everything we know into our formula for the product rule. And we get π‘₯ times two 𝑦 d𝑦 by dπ‘₯ plus 𝑦 squared times one, which is simply two π‘₯𝑦 d𝑦 by dπ‘₯ plus 𝑦 squared.

We know that the derivative of zero is zero. So when we differentiate our entire equation, we get 18π‘₯ squared plus two π‘₯𝑦 d𝑦 by dπ‘₯ plus 𝑦 squared minus 10𝑦 d𝑦 by dπ‘₯ equals zero. Remember, though, in order to find an equation of the tangent to the curve at our point, we need to find its slope. So we need an equation for d𝑦 by dπ‘₯. This time, that will be in terms of both π‘₯ and 𝑦. So what we’re going to do is factor d𝑦 by dπ‘₯ from two π‘₯𝑦 and negative 10𝑦. And then, we’re going to subtract 18π‘₯ squared and 𝑦 squared from both sides of our equation.

Next, we divide through by two π‘₯𝑦 minus 10𝑦. And we find d𝑦 by dπ‘₯ is equal to negative 18π‘₯ squared minus 𝑦 squared over two π‘₯𝑦 minus 10𝑦. It’s also important to realise that this is equivalent to the fraction 18π‘₯ squared plus 𝑦 squared over 10𝑦 minus two π‘₯𝑦. We could achieve this fraction either by multiplying our fraction by negative one on both the numerator and the denominator or by rearranging in a slightly different order a little bit earlier on.

It really doesn’t matter though. What we’re going to do is substitute π‘₯ equals two and 𝑦 equals four into our equation for the derivative. That’s negative 18 times two squared minus four squared over two times two times four minus 10 times four, which is equal to 11 over three. And that tells us the slope of the tangent to the curve at the point two, four. We can now substitute everything we know about our tangent into the equation of a straight line. 𝑦 minus 𝑦 one is 𝑦 minus four. π‘š, the slope, is eleven-thirds. And π‘₯ minus π‘₯ one is π‘₯ minus two.

Next, we’re going to distribute the parentheses on the right-hand side. We do this by multiplying both terms inside the bracket by eleven-thirds. So we get 𝑦 minus four equals eleven-thirds π‘₯ minus twenty-two thirds. We subtract eleven-thirds π‘₯ from both sides. And our final step will be to add this twenty-two thirds to both sides. Negative four can be written as negative 12 over three. Negative 12 over three plus 22 over three is 10 over three.

And so we found the equation of the tangent to the curve at the point two, four. It’s negative 11π‘₯ over three plus 𝑦 plus 10 over three equals zero.

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