京都大学情報学研究科知能情報学専攻 2021年2月実施基礎科目 F1-2

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Author

祭音Myyura (co-authored with GPT 5.6 SOL)

Given

z=2y-x^4y^2,\qquad y=\cos x,

compute $\partial z/\partial x$ and $\partial z/\partial y$ . Using these results, determine $dz/dx$ .

Evaluate

\int_{-1}^{0}\left(\frac{1}{1-x}+\sqrt{1-x}\right)\,dx.

Find the maximum rate of change of

f(x,y)=\sqrt{x^2+y^3}

at $(1,1)$ and the direction in which this maximum rate occurs.

已知 $z=2y-x^4y^2$ 、 $y=\cos x$ ，求 $\partial z/\partial x$ 、 $\partial z/\partial y$ ，并据此求 $dz/dx$ 。
计算

$\int_{-1}^{0}\left(\frac{1}{1-x}+\sqrt{1-x}\right)\,dx.$
求 $f(x,y)=\sqrt{x^2+y^3}$ 在 $(1,1)$ 处的最大变化率及其方向。

Treating $x$ and $y$ as independent variables,

\boxed{\frac{\partial z}{\partial x}=-4x^3y^2}, \qquad \boxed{\frac{\partial z}{\partial y}=2-2x^4y}.

Since $dy/dx=-\sin x$ , the chain rule gives

\begin{aligned} \frac{dz}{dx} &=\frac{\partial z}{\partial x} +\frac{\partial z}{\partial y}\frac{dy}{dx}\\ &=-4x^3\cos^2x-(2-2x^4\cos x)\sin x. \end{aligned}

Therefore,

\boxed{\frac{dz}{dx} =-4x^3\cos^2x-2\sin x+2x^4\sin x\cos x}.

An antiderivative is

-\log(1-x)-\frac23(1-x)^{3/2}.

Hence

\begin{aligned} \int_{-1}^{0}\left(\frac{1}{1-x}+\sqrt{1-x}\right)\,dx &=\log2+\frac23\left(2^{3/2}-1\right)\\ &=\boxed{\log2+\frac{4\sqrt2-2}{3}}. \end{aligned}

The gradient is

\nabla f(x,y)= \left( \frac{x}{\sqrt{x^2+y^3}}, \frac{3y^2}{2\sqrt{x^2+y^3}} \right).

At $(1,1)$ ,

\nabla f(1,1)= \left(\frac1{\sqrt2},\frac3{2\sqrt2}\right).

The maximum directional derivative is the gradient norm:

\boxed{\|\nabla f(1,1)\| =\sqrt{\frac12+\frac98} =\frac{\sqrt{26}}4}.

It occurs in the unit direction of the gradient,

\boxed{\boldsymbol{u} =\frac{\nabla f(1,1)}{\|\nabla f(1,1)\|} =\frac1{\sqrt{13}}(2,3)}.