You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Constant of integration problem
- Thread starter MrJoe2000
- Start date
D
Deleted member 4993
Guest
So I have an derivative of ds= tdt/(2t + 1)^1/2
when t=0, s=0
what is s when t=24?
I figured out my integral to be [(2t+1)^1/2]/2 + 1/(2(2t+1)^1/2) + C
The book answer is s= 54 but I cant seem to get there. Can anyone help?
Your integration is incorrect.
Please show your steps. The correct answer should have an exponent of (3/2). Try the integration again....
Hello, MrJoe2000!
I believe you integrated incorrectly.
\(\displaystyle \displaystyle\text{We want: }s \;=\;\int\frac{t\,dt}{\sqrt{2t+1}} \)
\(\displaystyle \text{Let }\,u \,=\,\sqrt{2t+1} \quad\Rightarrow\quad u^2 \,=\,2t+1 \quad\Rightarrow\quad t \,=\,\frac{u^2-1}{2} \quad\Rightarrow\quad dt \,=\,u\,du\)
\(\displaystyle \displaystyle\text{Substitute: }\:\int\frac{\left(\frac{u^2-1}{2}\right)(u\,du)}{u} \;=\;\frac{1}{2}\int(u^2-1)\,du\)
. . . . . . . . \(\displaystyle =\;\dfrac{1}{2}\left(\dfrac{u^3}{3} - u\right) + C \;=\;\dfrac{u}{6}(u^2-3) + C\)
\(\displaystyle \text{Back-substitute: }\:s \;=\;\dfrac{\sqrt{2t+1}}{6}\big[(2t+1) - 3\big] + C \;\;=\;\; \dfrac{\sqrt{2t+1}}{6}(2t-2) + C \)
. . . . . . . . . . . . . \(\displaystyle s \;=\;\dfrac{\sqrt{2t+1}(t-1)}{3} + C \)
\(\displaystyle \text{When }t = 0,\,s = 0\!:\;\;0 \;=\;\dfrac{\sqrt{1}(-1)}{3} + C \quad\Rightarrow\quad C \:=\:\frac{1}{3} \)
\(\displaystyle \text{Hence: }\:s \;=\;\frac{1}{2}(t-1)\sqrt{2t+1} + \frac{1}{3} \)
\(\displaystyle \text{When }t = 24\!:\;\;s \;=\;\frac{1}{3}(24-1)\sqrt{2(24)+1} + \frac{1}{3} \;=\;\frac{1}{3}(23)(7) + \frac{1}{3} \)
. . . . . . . . . . . . \(\displaystyle s \;=\;\frac{161}{3} + \frac{1}{3} \;=\;\frac{162}{3} \;=\;\boxed{54}\)
I believe you integrated incorrectly.
\(\displaystyle \text{I have: }\:ds\:=\:\dfrac{t\,dt}{(2t+1)^{\frac{1}{2}}}\)
\(\displaystyle \text{When }t=0,\,s=0\)
\(\displaystyle \text{What is }s\text{ when }t=24\,?\)
\(\displaystyle \displaystyle\text{We want: }s \;=\;\int\frac{t\,dt}{\sqrt{2t+1}} \)
\(\displaystyle \text{Let }\,u \,=\,\sqrt{2t+1} \quad\Rightarrow\quad u^2 \,=\,2t+1 \quad\Rightarrow\quad t \,=\,\frac{u^2-1}{2} \quad\Rightarrow\quad dt \,=\,u\,du\)
\(\displaystyle \displaystyle\text{Substitute: }\:\int\frac{\left(\frac{u^2-1}{2}\right)(u\,du)}{u} \;=\;\frac{1}{2}\int(u^2-1)\,du\)
. . . . . . . . \(\displaystyle =\;\dfrac{1}{2}\left(\dfrac{u^3}{3} - u\right) + C \;=\;\dfrac{u}{6}(u^2-3) + C\)
\(\displaystyle \text{Back-substitute: }\:s \;=\;\dfrac{\sqrt{2t+1}}{6}\big[(2t+1) - 3\big] + C \;\;=\;\; \dfrac{\sqrt{2t+1}}{6}(2t-2) + C \)
. . . . . . . . . . . . . \(\displaystyle s \;=\;\dfrac{\sqrt{2t+1}(t-1)}{3} + C \)
\(\displaystyle \text{When }t = 0,\,s = 0\!:\;\;0 \;=\;\dfrac{\sqrt{1}(-1)}{3} + C \quad\Rightarrow\quad C \:=\:\frac{1}{3} \)
\(\displaystyle \text{Hence: }\:s \;=\;\frac{1}{2}(t-1)\sqrt{2t+1} + \frac{1}{3} \)
\(\displaystyle \text{When }t = 24\!:\;\;s \;=\;\frac{1}{3}(24-1)\sqrt{2(24)+1} + \frac{1}{3} \;=\;\frac{1}{3}(23)(7) + \frac{1}{3} \)
. . . . . . . . . . . . \(\displaystyle s \;=\;\frac{161}{3} + \frac{1}{3} \;=\;\frac{162}{3} \;=\;\boxed{54}\)