CPM math solutions chapter 7 7.67-7.72

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genyjoannerubiera

02.03.2022

Mathematics

The equation of the tangent line is $y = x + 0.5$ ====== $x^2 + y^2 = (2x^2+2y^2-x)^2$ the derivative at a point is the slope of the tangent line to that point.to implicitly differentiate, we assume that $y$ is a function of $x$ (so we need to apply chain rule when differentiating y terms)differentiate both sides $\dfrac{\text{d}}{\text{d}x}\left(x^2 + y^2\right) = \dfrac{\text{d}}{\text{d}x}\left(2x^2+2y^2-x\right)^2$ for the left-hand side, use sum rule for differentiation: differentiate each term $\begin{aligned} \dfrac{\text{d}}{\text{d}x}\left(x^2\right) + \dfrac{\text{d}}{\text{d}x}\left(y^2\right) & = \dfrac{\text{d}}{\text{d}x}\left(2x^2+2y^2-x\right)^2 \\ 2x + 2y \tfrac{\text{d}y}{\text{d}x}& = \dfrac{\text{d}}{\text{d}x}\left(2x^2+2y^2-x\right)^2 \end{aligned}$ for the right-hand side, we need to use the chain rule.i will deal with right-hand side separately here, and then add it back to original equation after $\begin{aligned} \text{rhs} & = \dfrac{\text{d}}{\text{d}x}\left(2x^2+2y^2-x\right)^2 \\ & = 2\left(2x^2+2y^2-x\right)^1 \cdot \dfrac{\text{d}}{\text{d}x}\left(2x^2+2y^2-x\right) \\ & = 2\left(2x^2+2y^2-x\right)\left(4x+4y\tfrac{\text{d}y}{\text{d}x}-1\right) \end{aligned}$ let's just connect this together now $\begin{aligned} 2x + 2y \tfrac{\text{d}y}{\text{d}x} & = 2\left(2x^2+2y^2-x\right)\left(4x+4y\tfrac{\text{d}y}{\text{d}x}-1\right) \end{aligned}$ we do not want to simplify this. you can if you want, but it involves a lot of time and unnecessary algebra. you might make mistakes. i might go over the character limit for these answers (around 5000 characters). stuff like that. we do not need to solve for $\text{d}y/\text{d}x$ in terms of $x$ and $y$ when we want to find the derivative at a point for an implicit curve.since we are interested in the point $(0,0.5)$ , substitute in 0 for all $x$ and 0.5 for all $y$ . then solve for $\text{d}y/\text{d}x$ , which is the slope of the tangent line.that x-value of 0 will really us out. $\begin{aligned} 2(0) + 2(0.5) \tfrac{\text{d}y}{\text{d}x} & = 2\left(2(0)^2+2(0.5)^2-0\right)\left(4(0)+4(0.5)\tfrac{\text{d}y}{\text{d}x}-1\right) \\ 1 \tfrac{\text{d}y}{\text{d}x} & = 2\big(2(0.25)\big)\left(2\tfrac{\text{d}y}{\text{d}x}-1\right) \\ \tfrac{\text{d}y}{\text{d}x} & = 2(0.5)\left(2\tfrac{\text{d}y}{\text{d}x}-1\right) \\ \tfrac{\text{d}y}{\text{d}x} & = 2\tfrac{\text{d}y}{\text{d}x}-1 \\ 0 & = \tfrac{\text{d}y}{\text{d}x} - 1 \\ 1 & = \tfrac{\text{d}y}{\text{d}x} \end{aligned}$ since $\text{d}y/\text{d}x$ is 1 at the point $(0,0.5)$ , the slope of the tangent line at point $(0,0.5)$ is 1.therefore, the slope of the line using point-slope form is $\begin{aligned} y - y_1 & = m(x-x_1) \\ y - 0.5 & = 1(x-0) \\ y & = x + 0.5 \end{aligned}$

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