Problem 31 - Calculus of a Single Variable

Question

Find the slope of the tangent line to the graph at the given point. Bifolium: \(\left(x^{2}+y^{2}\right)^{2}=4 x^{2} y\) Point: \((1,1)\)

Step-by-Step Solution

Answer

The slope of the tangent line to the graph of the given bifolium equation at the point (1,1) is 0.

Step 1: Understand the equation and differentiate implicitly

First, differentiate each side of the given equation \(\left(x^{2}+y^{2}\right)^{2}=4 x^{2} y\) with respect to \(x\). Using the chain rule for the left side which gives \(4(x^2+y^2)(x+yy')\) and the product rule on the right side which gives \(4x^2y' + 8xy\). Very important is to keep track of \(y'\) since it represents the derivative of \(y\) with respect to \(x\).

Step 2: Calculate for \(y'\)

Now that we have the equation, isolate \(y'\) by subtracting \(8xy\) from both sides and then divide everything by \(4x^2 + 4(x^2+y^2)y\). This will give you the equation in terms of \(y'\) as follows: \( y' = \frac{4(x^{2}+y^{2})-8xy}{4x^{2}+4(x^2+y^2)y} \).

Step 3: Subtitute the point into the equation

Since we are interested in the slope of the tangent line at the point (1,1), substitute \(x = 1\) and \(y = 1\) into the equation from step 2. The solution to this will be the slope of the tangent line at the point (1,1).

Step 4: Simplify the expression

Upon substituting \(x = 1\) and \(y = 1\) in the equation, we then simplify the equation to get the final answer. This involves a bit of simple arithmetic and possibly some fractions which need to be reduced to simplest terms.