Hyperbolic functions and its tangent

[ultima9999]

Just want to check my answer.

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

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

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

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

$$\begin{align*} y - 2 = 0 (x - 0) \\ \Rightarrow y = 2 \end{align*}$$

Therefore, the equation of the tangent at point (0, 2) is $$y = 2$$

 

[benorin]

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

Just want to check my answer.

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

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

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

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

$$\begin{align*} y - 2 = 0 (x - 0) \\ \Rightarrow y = 2 \end{align*}$$

Therefore, the equation of the tangent at point (0, 2) is $$y = 2$$