Hyperbolic functions and its tangent (1 Viewer)

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Just wanna check my answer.

Find the equation of the tangent to the curve [tex]y^3 + x^2 \cosh y + \sinh^3 x = 8[/tex] at the point (0, 2)

I firstly found the derivative and the gradient of the curve at point (0, 2)

[tex]3y^2 \cdot \frac{dy}{dx} + x \cosh y + x^2 \sinh y \cdot \frac{dy}{dx} + 3 \sinh^3 x \cosh x = 0[/tex]
[tex]\Rightarrow \frac{dy}{dx} (3y^2 + x^2 \sinh y) = - x \cosh y - 3 \sinh^3 x \cosh x[/tex]
[tex]\Rightarrow \frac{dy}{dx} = \frac{- x \cosh y - 3 \sinh^3 x \cosh x}{3y^2 + x^2 \sinh y}[/tex]

Substitute [tex]x = 0\ \mbox{and}\ y = 2[/tex] into the equation and I get a gradient of 0 at point (0, 2).

[tex]\begin{align*}
y - 2 = 0 (x - 0) \\
\Rightarrow y = 2
\end{align*}[/tex]

Therefore, the equation of the tangent at point (0, 2) is [tex]y = 2[/tex]
 
Last edited:

benorin

Homework Helper
1,057
7
You missed a 2 coeff on the xcoshy term in the first step, I fixed it in the quote: you answer is correct however.

ultima9999 said:
Just wanna check my answer.

Find the equation of the tangent to the curve [tex]y^3 + x^2 \cosh y + \sinh^3 x = 8[/tex] at the point (0, 2)

I firstly found the derivative and the gradient of the curve at point (0, 2)

[tex]3y^2 \cdot \frac{dy}{dx} + 2x \cosh y + x^2 \sinh y \cdot \frac{dy}{dx} + 3 \sinh^3 x \cosh x = 0[/tex]
[tex]\Rightarrow \frac{dy}{dx} (3y^2 + x^2 \sinh y) = -2 x \cosh y - 3 \sinh^3 x \cosh x[/tex]
[tex]\Rightarrow \frac{dy}{dx} = \frac{-2 x \cosh y - 3 \sinh^3 x \cosh x}{3y^2 + x^2 \sinh y}[/tex]

Substitute [tex]x = 0\ \mbox{and}\ y = 2[/tex] into the equation and I get a gradient of 0 at point (0, 2).

[tex]\begin{align*}
y - 2 = 0 (x - 0) \\
\Rightarrow y = 2
\end{align*}[/tex]

Therefore, the equation of the tangent at point (0, 2) is [tex]y = 2[/tex]
 

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