방향도함수와 기울기벡터

(a) Let's use $\mathbf{v}$ to denote the bee's velocity vector at time $t=2$.

• The bee's direction of motion is tangent to the curve. That tangent is perpendicular to both the normal vector to $3z + x^2 + y^2 = 2$ at $(1,1,0)$, which is

$$\left< 2x\,,\,2y\,,\,3\right>\Big|_{(x,y,z)=(1,1,0)} = \left< 2\,,\,2\,,\,3\right>$$

and the normal vector to $z = x^2 - y^2$ at $(1,1,0)$, which is

$$\left< 2x\,,\,-2y\,,\,-1\right>\Big|_{(x,y,z)=(1,1,0)} = \left< 2\,,\,-2\,,\,-1\right>$$

So $\mathbf{v}$ has to be some constant times

$$\begin{aligned} \left< 2\,,\,2\,,\,3\right> \times \left< 2\,,\,-2\,,\,-1\right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 2 & 2 & 3 \\ 2 & -2 & -1 \end{matrix}\right] =\left< 4 \,,\, 8 \,,\, -8 \right> \end{aligned}$$

or, equivalently, some constant times $\left< 1 \,,\, 2 \,,\, -2 \right>$.

• Since the $z$–component of $\mathbf{v}$ has to be positive, $\mathbf{v}$ has to be a positive constant times $\left< -1 \,,\, -2 \,,\, 2 \right>$.
• Since the speed has to be $6$, $\mathbf{v}$ has to have length $6$.

As $|\left< -1 \,,\, -2 \,,\, 2 \right>|=3$

$$\mathbf{v} = 2\left< -1 \,,\, -2 \,,\, 2 \right> =\left< -2 \,,\, -4 \,,\, 4 \right>$$

(b) Solution 1: Suppose that the bee is at $\big(x(t),y(t),z(t)\big)$ at time $t$. Then the temperature that the bee feels at time $t$ is

$$\begin{aligned} T\big(x(t),y(t),z(t),t\big) = x(t) y(t) -3x(t) +2y(t) t +z(t) \end{aligned}$$

Then the rate of change of temperature (per unit time) felt by the bee at time $t=2$ is

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2} &=x'(2)y(2) + x(2)y'(2) -3x'(2) +2y'(2)2+2y(2) +z'(2) \\ \end{aligned}$$

Recalling that, at time $t=2$, the bee is at $(1,1,0)$ and has velocity $\left< -2 \,,\, -4 \,,\, 4 \right>$

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2} &=(-2)(1) + (1)(-4) -3(-2) +2(-4)2+2(1) +4 \\ &=-10 \end{aligned}$$

(b) Solution 2: Suppose that the bee is at $\big(x(t),y(t),z(t)\big)$ at time $t$. Then the temperature that the bee feels at time $t$ is

$$\begin{aligned} T\big(x(t),y(t),z(t),t\big) \end{aligned}$$

By the chain rule, the rate of change of temperature (per unit time) felt by the bee at time $t=2$ is

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2} &=\bigg[\frac{\partial T}{\partial x}\big(x(t),y(t),z(t),t\big)\,x'(t) +\frac{\partial T}{\partial y}\big(x(t),y(t),z(t),t\big)\,y'(t) \\& +\frac{\partial T}{\partial z}\big(x(t),y(t),z(t),t\big)\,z'(t) +\frac{\partial T}{\partial t}\big(x(t),y(t),z(t),t\big)\bigg]_{t=2} \end{aligned}$$

Recalling that $T = xy - 3x+2yt+z$, we have

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2} &=[y(2)-3]x'(2) + [x(2)+2\times 2]y'(2) +z'(2) +2y(2) \end{aligned}$$

Also recalling that, at time $t=2$, the bee is at $(1,1,0)$ and has velocity $\left< -2 \,,\, -4 \,,\, 4 \right>$

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2} &=[-2](-2) + [5](-4) +4 +2 \\ &=-10 \end{aligned}$$