Question

Find the equation of the tangent line to the graph of the implicitly defined function at the indicated points. As a visual aid, each function is graphed. \(\left(x^{2}+y^{2}-4\right)^{3}=108 y^{2}\) (a) At (0,4) . (b) At \((2,-\sqrt[4]{108})\)

Short answer

(a) The tangent line at (0,4) is \(y=4\). (b) The tangent line at \((2,-\sqrt[4]{108})\) is \(y=x-2-\sqrt[4]{108}\).

Step-by-step solution

Differentiate Implicitly

Start by differentiating both sides of the given equation with respect to \(x\). The given equation is \((x^{2}+y^{2}-4)^{3} = 108y^{2}\). Using implicit differentiation, differentiate the left side with respect to \(x\), applying the chain rule: \[3(x^{2} + y^{2} - 4)^{2}(2x + 2y\frac{dy}{dx})\]. Differentiating the right side with respect to \(x\), we apply the product rule \[216y\frac{dy}{dx}\].We now have the equation: \\[3(x^{2} + y^{2} - 4)^{2}(2x + 2y\frac{dy}{dx}) = 216y\frac{dy}{dx}\].

Isolate \(\frac{dy}{dx}\)

Rearrange the implicit differentiation equation to solve for \(\frac{dy}{dx}\). First, expand the left-hand side:\[6x(x^{2} + y^{2} - 4)^{2} + 6y(x^{2} + y^{2} - 4)^{2}\frac{dy}{dx} = 216y\frac{dy}{dx}\]. Subtract the \(6y(x^{2} + y^{2} - 4)^{2}\frac{dy}{dx}\) term from both sides:\[6y(x^{2} + y^{2} - 4)^{2}\frac{dy}{dx} - 216y\frac{dy}{dx} = -6x(x^{2} + y^{2} - 4)^{2}\]. Factor \(\frac{dy}{dx}\) out from the left side:\[\frac{dy}{dx}\left[ 6y(x^{2} + y^{2} - 4)^{2} - 216y \right] = -6x(x^{2} + y^{2} - 4)^{2}\].Finally, solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{-6x(x^{2} + y^{2} - 4)^{2}}{6y(x^{2} + y^{2} - 4)^{2} - 216y}\].

Evaluate \(\frac{dy}{dx}\) at (0, 4)

Substitute \(x = 0\) and \(y = 4\) into the derivative \(\frac{dy}{dx}\). Firstly, calculate \((x^2 + y^2 - 4)^2\) when \(x = 0\) and \(y = 4\):\[ (0^2 + 4^2 - 4) = 12 \]\[ 12^2 = 144 \].Now substitute these values into \(\frac{dy}{dx}\): \[\frac{dy}{dx} = \frac{-6 \cdot 0 \cdot 144}{6 \cdot 4 \cdot 144 - 216 \cdot 4}\]. Simplifying gives:\[\frac{dy}{dx} = \frac{0}{6 \cdot 4 \cdot 144 - 864}\].The derivative at \( (0, 4) \) is \(0\).

Formulate the Tangent Line at (0, 4)

Using the slope \(m = 0\) and point \((0, 4)\), the tangent line equation is \(y - y_1 = m(x-x_1)\). Therefore:\[y - 4 = 0(x - 0)\] which simplifies to:\[y = 4\].

Evaluate \(\frac{dy}{dx}\) at \((2, -\sqrt[4]{108})\)

Substitute \(x = 2\) and \(y = -\sqrt[4]{108}\) into \(\frac{dy}{dx}\). Calculate \((x^2 + y^2 - 4)^2\) with these values:\[ (2^2 + (-\sqrt[4]{108})^2 - 4)^2 = (4 + \sqrt{108} - 4)^2 = (\sqrt{108})^2 = 108 \].Substitute into \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{-6 \cdot 2 \cdot 108}{6(-\sqrt[4]{108})108 - 216(-\sqrt[4]{108})}\].Simplify the expression:\[\frac{dy}{dx} = \frac{-1296}{-1296}\], which gives \(\frac{dy}{dx} = 1\).

Formulate the Tangent Line at \((2, -\sqrt[4]{108})\)

With a slope of 1 and point \((2, -\sqrt[4]{108})\), use the point-slope form: \[ y - (-\sqrt[4]{108}) = 1(x - 2) \]This becomes:\[ y = x - 2 - \sqrt[4]{108}\].

Key concepts

Tangent Line Equation

When we talk about tangent lines on a graph, we're referring to lines that touch the curve at exactly one point, without crossing it. This tangent line tells us the slope or the direction of the curve precisely at that point.

To find the equation of a tangent line for any given function, we often use the derivative, which signifies the slope at any point on the curve. The derivative is denoted by \( \frac{dy}{dx} \), meaning the rate at which \( y \) changes concerning \( x \). Once you have the slope from the derivative, you can use the point-slope form of a line equation which is \( y - y_1 = m(x-x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is the specific point where you want to find the tangent.

For example, if the slope is zero, as it was in our exercise at point \((0,4)\), the tangent line is horizontal, thus \( y = 4 \). Understanding this equation is crucial in identifying how curves behave at specific points.

Implicitly Defined Functions

Implicit functions are those where \( y \) is not isolated on one side of the equation. This means you always have both \( x \) and \( y \) mixed together. Clearly distinguishing them can be tricky!

In these types of functions, you don't explicitly solve for \( y \) in terms of \( x \). The solution still exists, but somewhere within the confines of the equation.

• Implicit differentiation is the method used to find derivatives of these implicit functions.
• We differentiate both \( x \) and \( y \) simultaneously, and keep \( \frac{dy}{dx} \) in the equation until we isolate it.

In our exercise, we used implicit differentiation to handle the function \((x^{2}+y^{2}-4)^{3}=108 y^{2}\). By carefully applying rules of differentiation, we found the derivative, \( \frac{dy}{dx} \). Getting familiar with this technique is valuable for solving many real-world problems involving complex relationships.

Chain Rule

The chain rule is a fundamental tool in calculus used for differentiating compositions of functions. It helps when you have a function inside another function. Sometimes this can resemble a layered cake – you peel back one layer just to find another underneath!

Consider our equation: \( (x^{2}+y^{2}-4)^{3} \). The outside function is the cube of something, and the inside is \( x^{2} + y^{2} - 4 \). Applying the chain rule lets us differentiate each piece step by step. Here's the magic of the chain rule in action:

• Differentiate the outer layer while keeping the inner function untouched.
• Multiply by the derivative of the inner function.

In simpler terms, for \( (u(x))^{n} \), the derivative is \( nu^{n-1} \cdot \frac{du}{dx} \).

The power of the chain rule in our problem came through when we differentiated with respect to \( x \), showing how changes in \( x \) affect not just itself but the other intertwined elements. Mastering this tool enables deep dives into more complex, layered functions.