적분 전체 문제

$400$ft

$30$ft

$1.5/\ln(2) \approx 2.164$miles

$-1$ft

$128/5$ft

$-64$ft/s

$50$ft/s

$2$ft/s

$0$ft/s

(a) $$\begin{aligned} (ax)^2+(by)^2&=1\\ by&=\sqrt{1-(ax)^2}\\ y &= \frac{1}{b}\sqrt{1-(ax)^2} \end{aligned}$$ (b) The values of $x$ in the domain of the function above are those that satisfy $1-(ax)^2 \geq 0$. That is, $-\frac{1}{a}\leq x \leq \frac{1}{a}$. Therefore, the upper half of the ellipse has area

$$\begin{aligned} \displaystyle\frac{1}{b}&\displaystyle\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{1-(ax)^2}\ \mathrm{d}x \\ \text{The upper half of a circle has equation $y=\sqrt{r^2-x^2}$. } \\ &=\frac{1}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{a^2\left(\frac{1}{a^2}-x^2\right)}\ \mathrm{d}x\\ &=\frac{1}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}a\sqrt{\frac{1}{a^2}-x^2}\ \mathrm{d}x\\ &=\frac{a}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{\frac{1}{a^2}-x^2}\ \mathrm{d}x \end{aligned}$$ (c) The function $y=\sqrt{\dfrac{1}{a^2}-x^2}$ is the upper-half of the circle centred at the origin with radius $\dfrac{1}{a}$. So, the expression from (b) evaluates to $\left(\dfrac{a}{b}\right)\dfrac{\pi}{2a^2} = \dfrac{\pi}{2ab}$.

The expression from (b) was half of the ellipse, so the area of the ellipse is $\dfrac{\pi}{ab}$.

Remark: this was a slightly long-winded way of getting the result. The reasoning is basically this:

• The area of the unit circle $x^2+y^2=1$ is $\pi$ .
• The ellipse $(ax)^2+y^2=1$ is obtained by shrinking the unit circle horizontally by a factor of $a$. So, its area is $\dfrac{\pi}{a}$ .
• Further, the ellipse $(ax)^2+(by)^2=1$ is obtained from the previous ellipse by shrinking it vertically by a factor of $b$. So, its area is $\dfrac{\pi}{ab}$ .

The car stops when $v(t)=30-10t=0$, which occurs at time $t = 3$. The distance covered up to that time is

$$\begin{aligned} \int_0^3 v(t)\,\mathrm{d}t = (30t - 5t^2)\Big|_{0}^{3} = (90-45)-0 = 45\,\textrm{m}. \end{aligned}$$

(a) Setting $v(t) = 0$, we find $500 - 32t = 0$, and thus $t = \frac{500}{32} = \frac{125}{8} = 15.625$. This is the time that the rocket's velocity changes from positive to negative, and thus the point when the rocket's position changes from increasing to decreasing: it's when the rocket reaches its maximum height. (b) Since $v$ is a positive linear function that connects the points $(0,500)$ and $(15.625,0)$ on the interval $[0, 15.625]$, the region enclosed by the velocity function and the $t$-axis is a right triangle with vertical leg of length $500$ and horizontal leg of length $15.625$. The area of this triangle is $A = \frac{1}{2}bh = \frac{1}{2}(15.625)(500) = 3906.25$. Because this is the area under the velocity curve (where velocity is positive) this area represents the vertical distance traveled by the rocket on the time interval $[0, 15.625]$. (c) One function that has derivative $v(t) = s'(t) = 500-32t$ is $s(t) = 500t - 16t^2$. (d) Using our result in (c), $s(15.625) - s(0) = 500(15.625) - 16(15.625)^2 - (0 - 0) = 3906.25$. Thus, the change of the rocket's position on $[0,15.625]$ is $3906.25$ feet, which is also the distance the rocket traveled vertically on the time interval, since velocity is positive throughout. (e) We find that $s(5) - s(1) = 1616$, which tells us that the rocket rose $1616$ feet on the $4$-second interval from $t = 1$ to $t = 5$.

(a) We observe that $v(t) = 0$ at $t = 0$, $t = 1$, and $t = 3$, and $v(t)$ is positive on the intervals $0 < t < 1$ and $3 < t < 4$, while $v(t)$ is negative on $1 < t < 3$. The total change in position, $P$, of the object is thus given by $$P = \int_0^1 v(t) \, dt + \int_1^3 v(t) \, dt + \int_3^4 v(t) \, dt = \int_0^4 v(t) \, dt\,.$$ (b) Using computational technology, we use the midpoint rule to estimate the total change of position, $\int_0^4 v(t) \, dt$. We find that $M_{25} = 2.6496$, $M_{50} = 2.6624$, $M_{75} = 2.66477$, and $M_{100} = 2.6656$. Since it appears that neither of the first two decimal places are changing as we increase the number of subintervals, it looks like $P = \int_0^4 v(t) \, dt \approx 2.665$ is accurate to two decimal places. (c) We noted in (a) where $v(t)$ is positive and negative. In particular, on $1 < t < 3$, $v(t) < 0$, and thus $\int_1^3 v(t) \, dt$ will give us a negative change in position; the distance traveled on this interval is therefore $-\int_1^3 v(t) \, dt$, and hence the total distance traveled, $D$, on $[0,4]$ is $$D = \int_0^1 v(t) \, dt - \int_1^3 v(t) \, dt + \int_3^4 v(t) \, dt\,.$$ (d) To use Riemann sums to estimate $D$, we have to estimate each of the three integrals individually. We find that

• $\int_0^1 v(t) \, dt \approx M_{100} = 0.41669$,
• $\int_1^3 v(t) \, dt \approx M_{100} = -2.6668$, and
• $\int_1^4 v(t) \, dt \approx M_{100} = 4.91667$ From experimenting with different $n$-values in $M_n$, it is apparent that each of these individual estimates is accurate to at least $3$ decimal places, and thus we expect their sum to be accurate to at least $2$. Hence, $$D = \int_0^1 v(t) \, dt - \int_1^3 v(t) \, dt + \int_3^4 v(t) \, dt \approx 0.41669 - (-2.6668) + 4.91667 = 8.00016\,.$$ Therefore $D \approx 8.00016$ is an estimate of distance traveled accurate to at least $2$ decimal places. (e) We know that the average velocity on $[0,4]$ is given by the integral expression $$v_{\operatorname{AVG} [0,4]} = \frac{1}{4-0} \int_0^4 v(t) \, dt\,.$$ We estimated the definite integral in (b), so using that result, $$v_{\operatorname{AVG} [0,4]} \approx \frac{1}{4-0} 2.665 = 0.66625$$ feet per second is the object's average velocity, accurate to at least two decimal places. It makes sense that the average velocity is positive since the total change in position is positive.