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

Find the equation of the tangent to the curve 9π‘₯Β³ βˆ’ 6π‘₯Β² + 6π‘₯ βˆ’ 𝑦² βˆ’ 𝑦 + 2 = 0 at the point (0, 1).

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Video Transcript

Find the equation of the tangent to the curve nine π‘₯ cubed minus six π‘₯ squared plus six π‘₯ minus 𝑦 squared minus 𝑦 plus two equals zero at the point zero, one.

We want to find the equation of the tangent line to the curve, which requires knowing its slope which we find by differentiating the equation given. And knowing the slope of the tangent line isn’t enough to know what its equation is. But we’re given the point at which the line is tangent to the curve. And so the tangent line passes through this point. And so having found the slope of the tangent line, we can use this point to find the point-slope equation of the tangent line.

Okay, first, let’s find the slope. We do this by differentiating the equation of the curve implicitly with respect to π‘₯. In other words, we differentiate both sides of the equation with respect to π‘₯. Finding the derivatives of the first three terms is straightforward. We use the fact that the derivative of π‘Žπ‘₯ to the 𝑛 with respect to π‘₯ is π‘Ž times 𝑛 times π‘₯ to the 𝑛 minus one. And so, we get 27π‘₯ squared minus 12π‘₯ plus six.

The derivative of 𝑦 squared with respect to π‘₯ however is a bit more tricky. We have to apply the chain rule. Now taking 𝑓 to be 𝑦 squared, we get 𝑑 by 𝑑𝑦 of 𝑦 squared times 𝑑𝑦 by 𝑑π‘₯ and 𝑑 by 𝑑𝑦 of 𝑦 squared is two 𝑦. So putting it altogether, we get two 𝑦 times 𝑑𝑦 by 𝑑π‘₯. The derivative of 𝑦 with respect to π‘₯ is much easier. It’s just 𝑑𝑦 by 𝑑π‘₯. And the derivative of two with respect to π‘₯ is zero as is the derivative of zero with respect to π‘₯.

Now, we have an equation involving 𝑑𝑦 by 𝑑π‘₯. We can rearrange it to make 𝑑𝑦 by 𝑑π‘₯ the subject. We do that by taking out the common factor of 𝑑𝑦 by 𝑑π‘₯ from the two terms which contain it, remembering here to be slightly careful about the minus signs. Now, we want to subtract the other terms to get them on the other side. And finally, we divide by minus two 𝑦 plus one. And we get 𝑑𝑦 by 𝑑π‘₯ in terms of π‘₯ and 𝑦.

We’re looking for the equation of the tangent at the point zero, one. So the value of π‘₯ is zero and the value of 𝑦 is one. And substituting, we get the following. Simplifying, we get negative six over negative three, which is two. So the slope of our tangent is two. Having found the slope of the tangent, we use it and the fact that the tangent goes to the point zero, one to find the equation of the tangent.

Here is the general equation of the straight line, which goes through the point π‘₯ nought, 𝑦 nought and has slope π‘š. Substituting our values, we get that 𝑦 minus one is equal to two times π‘₯ minus zero. So 𝑦 minus one equals two π‘₯ or rearranging slightly, 𝑦 minus two π‘₯ minus one equals zero.

As usual, we find the equation of the tangent to the curve by first differentiating the equation of the curve to find the slope of the tangent. And then, we use this value of the slope along with the point of contact of the curve with the tangent in the point-slope equation of a line.

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