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

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

Guest Jan 7, 2015

#1
+19054
+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
+19054
+10
$$\small{\text{  \begin{array}{rcl} \dfrac{\ ds}{\ dt} &=& t(3t-2) = 3t^2-2t \\\\ \ ds &=& (3t^2-2t) \ dt \quad | \quad \int\\ \\ \int{\ ds} = s &=& \int{ (3t^2-2t) \ dt }\\ \\ s &=& 3\int{t^2\ dt }-2\int{t\ dt } \\ \\ s &=& 3 \frac{t^3}{3} -2\frac{t^2}{2} +c\\ \\ s &=& t^3 - t^2 +c\\ \\ \end{array}  }}$$
$$\small{\text{  t = 1  and  s= 0  so we find c :  \begin{array}{rcl} 0 & = & 1^3 - 1^2 + c \\ 0 & = & 0 + c \\ 0 & = & c \end{Array}  }}\\ \small{\text{  \boxed{s = t^3-t^2 }  }}$$