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sharathe
Group Title
Can someone help me with this initial value problem? d^2y over dt^2 = 2/t^3. Given is: dr/dt = t=1 and that whole thing = 1. Another Given is: r(1) = 1. I have no idea how to go about this problem after finding the antiderivative for the d^2r over dx^2.
 2 years ago
 2 years ago
sharathe Group Title
Can someone help me with this initial value problem? d^2y over dt^2 = 2/t^3. Given is: dr/dt = t=1 and that whole thing = 1. Another Given is: r(1) = 1. I have no idea how to go about this problem after finding the antiderivative for the d^2r over dx^2.
 2 years ago
 2 years ago

This Question is Open

freckles Group TitleBest ResponseYou've already chosen the best response.1
so you never meant to say anything about y right?
 2 years ago

sharathe Group TitleBest ResponseYou've already chosen the best response.0
Sorry meant r not y.
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
\[r''=2t^{3}\]
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
So you said you integrated both sides: \[r'=2 \cdot \frac{t^{3+1}}{3+1}+C_1=2 \cdot \frac{t^{2}}{2}+C_1=1 t^{2}+C_1\] => \[r'=\frac{1}{t^2}+C_1\]
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
So we are given r'(1)=1?
 2 years ago

sharathe Group TitleBest ResponseYou've already chosen the best response.0
Just r(1)=1
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
dr/dt = t=1? r'=t=1 I look at that is when t=1 ,r '=1
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
\[1=\frac{1}{1^2}+C_1 \] Solve for C_1
 2 years ago

sharathe Group TitleBest ResponseYou've already chosen the best response.0
where does the r' come from?
 2 years ago

sharathe Group TitleBest ResponseYou've already chosen the best response.0
Oh! so another way to look at d^2r/dt^2 is r''?
 2 years ago

sharathe Group TitleBest ResponseYou've already chosen the best response.0
Okay! Also just to clarify, when substituting t and r, just plug in one in newly acquired derivative where r and t are respectively? If r(1)=1 and t(1)=2, then plug in respectively and then ... what would happen?
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
well if you are given r'(a)=b then you use the function for r' If t=a then r'=b  and if you are given r(c)=d then use the function r If t=c, then r=d You will use both of this conditions to find your constants
 2 years ago

sharathe Group TitleBest ResponseYou've already chosen the best response.0
So, since r(1)=1, then t(1)=2, then r of t = 1? Somehow I confused myself...
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
t(1)=2 makes no sense
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
I'm sorry to be so blunt
 2 years ago

sharathe Group TitleBest ResponseYou've already chosen the best response.0
No, it makes no sense to me either.
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
You are given r'(1)=1 and r(1)=1
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
dw:1332451923490:dw
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
r'(1)=1 says when t=1 r'=1
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
\[1=\frac{1}{1^2}+C_1 \] The reason why I wrote this
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
I replaced t with 1 and I replaced r' with 1
 2 years ago

sharathe Group TitleBest ResponseYou've already chosen the best response.0
That makes sense...what happens to the 2 of t(1)=2 tho? how would I solve for C then? Or is it just replace r' with 1 and t with 1?
 2 years ago

sharathe Group TitleBest ResponseYou've already chosen the best response.0
Sorry for all these questions...
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
I don;t understand where you are seeing t(1)=2?
 2 years ago

sharathe Group TitleBest ResponseYou've already chosen the best response.0
I added it after getting the orignal problem in the "what happens after" part...cause the next problem was basically this problem, but instead of t(1)=1, it was t(2)=1...Sorry if this was all confusing! I don't quite get how to solve for C when the r' and t equal different numbers but have the same number plugged in...
 2 years ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.2
@sharathe you're confused between the input t = 1, output r' ( 1) = 1 From these input and output => constant C1 = 2
 2 years ago
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