91影视

Slope is given by $\frac{\mathbf{dy}}{\mathbf{dx}}$

So, slope of the solution curve at $\left(\mathbf{1}\mathbf{,}\frac{蟺}{\mathbf{2}}\right)$is

$$\begin{gathered} \frac{\mathbf{dy}}{\mathbf{dx}}\left|\left(\mathbf{1}\mathbf{,}\frac{蟺}{\mathbf{2}}\right)\right.\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{sin}\left(\mathbf{1}\mathbf{,}\frac{蟺}{\mathbf{2}}\right) \\ \mathbf{=}\mathbf{1}\mathbf{+}\mathbf{1} \\ \mathbf{=}\mathbf{2} \end{gathered}$$

Hence, the slope at the point is 2.

Since, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{sin}鈥�\mathbf{y}$ for all $\mathbf{y}鈭�\mathrm{鈩�}$

Then, $\mathbf{x}\mathbf{+}\mathbf{sin}\mathbf{y}\mathbf{>}\mathbf{1}$, for all $\mathbf{x}\mathbf{>}\mathbf{1}$.

i.e., $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{>}\mathbf{1}\mathbf{>}\mathbf{0}$, for all $\mathbf{x}\mathbf{>}\mathbf{1}$, $\mathbf{y}鈭�\mathrm{鈩�}$.

Hence, from first derivative test, every solution curve is increasing for $\mathbf{x}\mathbf{>}\mathbf{1}$.

Here

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{sin}鈥�\mathbf{y}$

Differentiate both sides with respect to x

$$\begin{gathered} \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{cos}\mathbf{y}\left(\frac{\mathbf{dy}}{\mathbf{dx}}\right) \\ \mathbf{=}\mathbf{1}\mathbf{+}\mathbf{cos}\mathbf{y}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{sin}\mathbf{y}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{cos}\mathbf{y}\mathbf{+}\mathbf{sin}\mathbf{y}\mathbf{cos}\mathbf{y} \\ \mathbf{=}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{cos}\mathbf{y}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{sin}\mathbf{2}\mathbf{y} \end{gathered}$$

So, the second derivative of every solution satisfies the given equation.

Since, $\mathbf{x}\mathbf{+}\mathbf{sin}鈥�鈥�\mathbf{y}\mathbf{=}\mathbf{0}$at $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$

we get, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{0}$at $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$

thus, $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$ is a critical point.

Also, from part (c) ,$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{cos}\mathbf{y}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{sin}\mathbf{2}\mathbf{y}$

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{>}\mathbf{0}$at $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$

From second derivative test, $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$ is a point of relative minimum.

Therefore, the curve has a minimum at $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$