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Chapter 2: Problem 31

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: \((1,1)\)

Short Answer

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

Step by step solution

01

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 \(x\).
02

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: \( y' = \frac{4(x^{2}+y^{2})-8xy}{4x^{2}+4(x^2+y^2)y} \).
03

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).
04

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.

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