How do I find this tangent line?

Calculus Level 4

Let y = y ( x ) y = y(x) be the function implicitly determined by the equation x 2 + 2 x y + 3 ln y = e y . x^2 +2xy + 3 \ln{y} = e^y .

Find the slope (angular coefficient) of the line tangent to y ( x ) y(x) on y = 1 y = 1 , located at the first quadrant of the Cartesian plane.

Give your answer to 1 decimal place.


The answer is -1.8.

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1 solution

Letting y = 1 y=1 on the equation, we get

x 2 + 2 x = e x 2 + 2 x + 1 = e + 1 x + 1 = ± e + 1 x = 1 + e + 1 x^2 + 2x = e \Leftrightarrow x^2 + 2x + 1 = e + 1 \Leftrightarrow x+1 = \pm \sqrt{e+1} \Leftrightarrow \boxed{x = -1 + \sqrt{e+1}}

since we ought to have x > 0 , y > 0 x>0, \: y>0 so the point is located on the first quadrant of the Cartesian plane.

Using implicit derivation techniques on the displayed equation, we get

2 x + 2 y + 2 x d y d x + 3 y d y d x = e y d y d x 2x + 2y + 2x \cdot \dfrac{dy}{dx} + \frac{3}{y} \cdot \dfrac{dy}{dx} = e^y \cdot \dfrac{dy}{dx} \Leftrightarrow d y d x = 2 x + 2 y e y 2 x 3 y \Leftrightarrow \dfrac{dy}{dx} = \dfrac{2x+2y}{e^y -2x - \frac{3}{y}}

Therefore, d y d x ( 1 + e + 1 , 1 ) = 2 ( 1 + e + 1 ) + 2 e 2 ( 1 + e + 1 ) 3 = 2 e + 1 e 2 e + 1 1 1.8 \dfrac{dy}{dx} (-1 + \sqrt{e+1}, \; 1) = \dfrac{2 \cdot (-1 + \sqrt{e+1})+2}{e -2 \cdot (-1 + \sqrt{e+1}) - 3} = \boxed{\dfrac{2\sqrt{e+1}}{e - 2\sqrt{e+1} - 1} \approx -1.8}

I've never heard of the term "angular coefficient of a line". I would recommend using the more common term of slope.

Jon Haussmann - 3 years, 9 months ago

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It is incredibly common in Brazilian education! Sorry if you were not familiarized with it.

Guilherme Dela Corte - 3 years, 8 months ago

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Ok, that's fair. All I can say is that it is not a common term in American mathematics.

Jon Haussmann - 3 years, 8 months ago

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