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Write an integral that quantifies the increase in the surface area of a sphere as its radius doubles from R unit to 2 R units and evaluate the integral.
$12\pi {R}^{2}=8\pi {\displaystyle {\int}_{R}^{2R}rdr}$
Write an integral that quantifies the increase in the volume of a sphere as its radius doubles from R unit to 2 R units and evaluate the integral.
Suppose that a particle moves along a straight line with velocity $v\left(t\right)=4-2t,$ where $0\le t\le 2$ (in meters per second). Find the displacement at time t and the total distance traveled up to $t=2.$
$d\left(t\right)={\displaystyle {\int}_{0}^{t}v\left(s\right)ds=4t-{t}^{2}}.$ The total distance is $d(2)=4\phantom{\rule{0.2em}{0ex}}\text{m}\text{.}$
Suppose that a particle moves along a straight line with velocity defined by $v\left(t\right)={t}^{2}-3t-18,$ where $0\le t\le 6$ (in meters per second). Find the displacement at time t and the total distance traveled up to $t=6.$
Suppose that a particle moves along a straight line with velocity defined by $v\left(t\right)=\left|2t-6\right|,$ where $0\le t\le 6$ (in meters per second). Find the displacement at time t and the total distance traveled up to $t=6.$
$d\left(t\right)={\displaystyle {\int}_{0}^{t}v\left(s\right)ds}.$ For $t<3,d\left(t\right)={\displaystyle {\int}_{0}^{t}\left(6-2t\right)dt=6t-{t}^{2}}.$ For $t>3,d(t)=d(3)+{\displaystyle {\int}_{3}^{t}(2t-6)dt}=9+({t}^{2}-6t).$ The total distance is $d(6)=9\phantom{\rule{0.2em}{0ex}}\text{m}\text{.}$
Suppose that a particle moves along a straight line with acceleration defined by $a\left(t\right)=t-3,$ where $0\le t\le 6$ (in meters per second). Find the velocity and displacement at time t and the total distance traveled up to $t=6$ if $v\left(0\right)=3$ and $d\left(0\right)=0.$
A ball is thrown upward from a height of 1.5 m at an initial speed of 40 m/sec. Acceleration resulting from gravity is −9.8 m/sec ^{2} . Neglecting air resistance, solve for the velocity $v\left(t\right)$ and the height $h\left(t\right)$ of the ball t seconds after it is thrown and before it returns to the ground.
$v\left(t\right)=40-9.8t;h\left(t\right)=1.5+40t-4.9{t}^{2}$ m/s
A ball is thrown upward from a height of 3 m at an initial speed of 60 m/sec. Acceleration resulting from gravity is −9.8 m/sec ^{2} . Neglecting air resistance, solve for the velocity $v\left(t\right)$ and the height $h\left(t\right)$ of the ball t seconds after it is thrown and before it returns to the ground.
The area $A\left(t\right)$ of a circular shape is growing at a constant rate. If the area increases from 4 π units to 9 π units between times $t=2$ and $t=3,$ find the net change in the radius during that time.
The net increase is 1 unit.
A spherical balloon is being inflated at a constant rate. If the volume of the balloon changes from 36 π in. ^{3} to 288 π in. ^{3} between time $t=30$ and $t=60$ seconds, find the net change in the radius of the balloon during that time.
Water flows into a conical tank with cross-sectional area πx ^{2} at height x and volume $\frac{\pi {x}^{3}}{3}$ up to height x . If water flows into the tank at a rate of 1 m ^{3} /min, find the height of water in the tank after 5 min. Find the change in height between 5 min and 10 min.
At $t=5,$ the height of water is $x={\left(\frac{15}{\pi}\right)}^{1\text{/}3}\phantom{\rule{0.2em}{0ex}}\text{m}\text{.}.$ The net change in height from $t=5$ to $t=10$ is ${\left(\frac{30}{\pi}\right)}^{1\text{/}3}-{\left(\frac{15}{\pi}\right)}^{1\text{/}3}$ m.
A horizontal cylindrical tank has cross-sectional area $A\left(x\right)=4\left(6x-{x}^{2}\right){m}^{2}$ at height x meters above the bottom when $x\le 3.$
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