선형근사와 미분소

$dV = \pm 0.157$

(a) $\pm$12.8 feet (b) $\pm$32 feet

$\pm15\pi/8 \approx \pm5.89$in$^2$

Let $f(x)$ be the number of problems finished after $x$ minutes of work. The question tells us that $5\Delta y \approx \Delta x$. So, if $\Delta x = 15$, $\Delta y \approx 3$. That is, in 15 minutes more, you will finish about 3 more problems.

Remark: the math behind this problem is intended to be easy! Looking at symbols like $\Delta y$ and $f\left(x+\Delta x\right)$ can be confusing, but the basic idea is pretty simple.

(a)

When $x$ is near 4, the linear approximation of $\Delta y$ is

$$\begin{aligned} \Delta y &\approx s'(4)\Delta x \\ \text{From the problem, $\Delta x= 5-4=1$. Differentiating,} \\ s'(t)&=\sqrt{19.6}\cdot\frac{1}{2\sqrt{x}}=\sqrt{\frac{4.9}{x}},\text{ so }\\ s'(4)&=\sqrt{\frac{4.9}{4}} \\ \text{The linear approximation gives us} \\ \Delta y& \approx \sqrt{\frac{4.9}{4}}(1)\approx1.1 \end{aligned}$$

So moving from 4 metres to 5 metres increases the speed with which you hit the water by about 1.1 metres per second.

(b) Again, we'll use the linear approximation

$$\begin{aligned} \Delta y &\approx s'(a)\Delta x\\ &=\sqrt{\frac{4.9}{a}}\cdot\Delta x \end{aligned}$$

The difference in height between the first two jumps and between the last two jumps is the same, $\Delta x$. The initial height of the first jump is smaller than the initial height of the second jump. So, the value corresponding to $a$ is smaller for the first jump than for the second. Therefore, $\Delta y$ is larger between the first two jumps than it is between the last two jumps. So, the increase in speed from the first jump to the second is larger than the increase in speed from the second jump to the third.

To put that more symbolically, the change in terminal speed between the first two jumps is

$$\begin{aligned} \Delta y &\approx \sqrt{\frac{4.9}{x}}\cdot\Delta x \\ \text{while the change in terminal speed between the next two jumps is} \\ \Delta y &\approx \sqrt{\frac{4.9}{x+\Delta x}}\cdot\Delta x \end{aligned}$$

Since $\sqrt{\dfrac{4.9}{x}}\cdot\Delta x> \sqrt{\dfrac{4.9}{x+\Delta x}}\cdot\Delta x$ (when $x$ and $\Delta x$ are positive), the change in terminal speed is greater between the first two jumps than between the next two jumps.

(a) The linearization $L$ of $s$ at $t=a$ is $L(t) = s(a) + s'(a)(t-a)$. With the information we have we can find the linearization of $s$ at $t=9$, using the facts that $s(9) = 4$ and $s'(9) = -1.2$. Thus, the linearization of $s$ at $t=9$ is $L(t) = 4 - 1.2(t-9)$. One use of the linearization is that for $t$ close to $9$, $L(t)$ can provide a good approximation to $s(t)$. So $$s(9.34) \approx L(9.34) = 4-1.2(0.34) = 3.592\,.$$ (b) Since $s''(4) = 0.08$ is positive, we know that the graph of $s$ is concave up at $t=4$. This means that the linearization of $s$ at $t=4$ lies below the graph of $s$ and so $L(9.34)$ is an underestimate to $s(9.34)$. (c) Given that $s'(4)$ is negative, the object is moving back toward its original position. The fact that $s''(4)$ is positive means that the velocity of the object is increasing at $t=4$. This means that the object is actually slowing down as it moves toward toward its starting position at $t=4$.