Q. 83 - Calculus

Question

Your local weatherman predicts that there will be a rainstorm this afternoon for three hours and that the rate of rainfall will be $r\left(t\right)=0.6t^2\sqrt{3-t}$ inches per hour from $t=0 \mathrm{to} t=3$, as shown here. Find the total amount of rainfall predicted for the storm.

Step-by-Step Solution

Answer

The total amount of rainfall predicted for the storm is $4.254 \mathrm{inches}$.

Step 1. Given Information

Your local weatherman predicts that there will be a rainstorm this  afternoon for three hours and that the rate of rainfall will be $r\left(t\right)=0.6t^2\sqrt{3-t}$ inches per hour from $t=0 \mathrm{to} t=3$, as shown here. Find the total amount of rainfall predicted for the storm.

Step 2. The rate of rainfall will be r ( t ) = 0 . 6 t 2 3 − t inches per hour from t = 0   to   t = 3 .

$$\begin{gathered} {\int }_0^{3}r\left(t\right)\mathrm{d}t={\int }_0^{3}0.6t^2\sqrt{3-t}\mathrm{d}t \\ {\int }_0^{3}r\left(t\right)\mathrm{d}t=0.6{\int }_0^3{t}^2\sqrt{3-t}\mathrm{d}t \end{gathered}$$

Step 3. Using the substitution method.

$$\begin{gathered} u=3-t \\ \frac{du}{dt}=-1 \\ -du=dt \end{gathered}$$

In this question in which integration by substitution works even though the integrand is not at all in the form $f\text{'}\left(u\right(x\left)\right)u\text{'}\left(x\right)$. A clever change of variables will allow us to rewrite the integral so that it can be algebraically simplified.

$t=3-u$

Now the limit is

$$\begin{gathered} t=0 t=3 \\ u=3-t u=3-3 \\ u=3-0 u=0 \\ u=3 u=0 \end{gathered}$$

Step 4. This substitution changes the integral into

$$\begin{gathered} {\int }_0^{3}r\left(t\right)\mathrm{d}t=-0.6{\int }_3^0{(3-u)}^2\sqrt{u}\mathrm{du} \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6{\int }_3^0\left\{{{\left(3\right)}^2-2\times 3\times \mathrm{u}+\left(\mathrm{u}\right)}^2\right\}{\mathrm{u}}^{1/2}\mathrm{du} \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6{\int }_3^0{(9-6\mathrm{u}+\mathrm{u}}^2\left){\mathrm{u}}^{1/2}\mathrm{du} \\ {\int }_0^3\mathrm{r}\right(\mathrm{t})\mathrm{dt}=-0.6{\int }_3^0{(9{\mathrm{u}}^{1/2}-6\mathrm{u}·{\mathrm{u}}^{1/2}+\mathrm{u}}^2·{\mathrm{u}}^{1/2})\mathrm{du} \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6{\int }_3^0{(9{\mathrm{u}}^{1/2}-6{\mathrm{u}}^{3/2}+\mathrm{u}}^{5/2})\mathrm{du} \end{gathered}$$

Step 5. After simplifying.

$$\begin{gathered} {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6\left[9{\int }_3^0{\mathrm{u}}^{1/2}\mathrm{du}-6{\int }_3^0{\mathrm{u}}^{3/2}\mathrm{du}+{\int }_3^0{\mathrm{u}}^{5/2}\mathrm{du}\right] \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6{\left[9\left\{\frac{{\mathrm{u}}^{1/2+1}}{1/2+1}\right\}-6\left\{\frac{{\mathrm{u}}^{3/2+1}}{3/2+1}\right\}+\left\{\frac{{\mathrm{u}}^{5/2+1}}{5/2+1}\right\}\right]}_3^0 \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6{\left[9\left\{\frac{{\mathrm{u}}^{3/2}}{3/2}\right\}-6\left\{\frac{{\mathrm{u}}^{5/2}}{5/2}\right\}+\left\{\frac{{\mathrm{u}}^{7/2}}{7/2}\right\}\right]}_3^0 \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6{\left[9·\frac{2}{3}{\mathrm{u}}^{3/2}-6·\frac{2}{5}{\mathrm{u}}^{5/2}+\frac{2}{7}{\mathrm{u}}^{7/2}\right]}_3^0 \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6{\left[6{\mathrm{u}}^{3/2}-\frac{12}{5}{\mathrm{u}}^{5/2}+\frac{2}{7}{\mathrm{u}}^{7/2}\right]}_3^0 \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6{\left[6{(3-t)}^{3/2}-\frac{12}{5}{(3-t)}^{5/2}+\frac{2}{7}{(3-t)}^{7/2}\right]}_3^0 \end{gathered}$$

Step 6. Now finding the total amount of rainfall predicted for the storm.

$$\begin{gathered} {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6\left[6\left\{{(3-3)}^{3/2}-{(3-0)}^{3/2}\right\}-\frac{12}{5}\left\{{(3-3)}^{5/2}-{(3-0)}^{5/2}\right\}+\frac{2}{7}\left\{{(3-3)}^{7/2}-{(3-0)}^{7/2}\right\}\right] \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6\left[6\left\{0-{\left(3\right)}^{3/2}\right\}-\frac{12}{5}\left\{0-{\left(3\right)}^{5/2}\right\}+\frac{2}{7}\left\{0-{\left(3\right)}^{7/2}\right\}\right] \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6\left[-6{\left(3\right)}^{3/2}+\frac{12}{5}{\left(3\right)}^{5/2}-\frac{2}{7}{\left(3\right)}^{7/2}\right] \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6\left[-6\sqrt{3^3}+\frac{12}{5}\sqrt{3^5}-\frac{2}{7}\sqrt{3^7}\right] \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6\left[-6\sqrt{27}+\frac{12}{5}\sqrt{243}-\frac{2}{7}\sqrt{2187}\right] \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6\left[-6\times 5.19+\frac{12}{5}\times 15.59-\frac{2}{7}\times 46.8\right] \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6\left[-31.14+37.42-13.37\right] \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=-0.6\times (-7.09) \\ {\int }_0^3\mathrm{r}\left(\mathrm{t}\right)\mathrm{dt}=4.254 \end{gathered}$$