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ds/dt=t(3t-2) and s=0 when t=1

Guest Jan 7, 2015

Best Answer 

 #1
avatar+18829 
+10

ds/dt=t(3t-2) and s=0 when t=1

$$\small{\text{
$
\begin{array}{rcl}
\dfrac{\ ds}{\ dt} &=& t(3t-2) = 3t^2-2t \\\\
\ ds &=& (3t^2-2t) \ dt \quad | \quad \int\\ \\
\int{\ ds} = s &=& \int{ (3t^2-2t) \ dt }\\ \\
s &=& 3\int{t^2\ dt }-2\int{t\ dt } \\ \\
s &=& 3 \frac{t^3}{3} -2\frac{t^2}{2} +c\\ \\
s &=& t^3 - t^2 +c\\ \\
\end{array}
$
}}$$

$$\small{\text{
$
t = 1 $ and $ s= 0 $ so we find c :
$
\begin{array}{rcl}
0 & = & 1^3 - 1^2 + c \\
0 & = & 0 + c \\
0 & = & c
\end{Array}
$
}}$\\$
\small{\text{
$
\boxed{s = t^3-t^2 }
$
}}$$

heureka  Jan 7, 2015
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1+0 Answers

 #1
avatar+18829 
+10
Best Answer

ds/dt=t(3t-2) and s=0 when t=1

$$\small{\text{
$
\begin{array}{rcl}
\dfrac{\ ds}{\ dt} &=& t(3t-2) = 3t^2-2t \\\\
\ ds &=& (3t^2-2t) \ dt \quad | \quad \int\\ \\
\int{\ ds} = s &=& \int{ (3t^2-2t) \ dt }\\ \\
s &=& 3\int{t^2\ dt }-2\int{t\ dt } \\ \\
s &=& 3 \frac{t^3}{3} -2\frac{t^2}{2} +c\\ \\
s &=& t^3 - t^2 +c\\ \\
\end{array}
$
}}$$

$$\small{\text{
$
t = 1 $ and $ s= 0 $ so we find c :
$
\begin{array}{rcl}
0 & = & 1^3 - 1^2 + c \\
0 & = & 0 + c \\
0 & = & c
\end{Array}
$
}}$\\$
\small{\text{
$
\boxed{s = t^3-t^2 }
$
}}$$

heureka  Jan 7, 2015

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