Differentiate x2=1+t2 find acceleration
Answers (2)
10th Jul, 2019
Hi V k,
Here is the solution for your query
x=(1+t^2)^1/2
Now we need to differentiate this twice, that is double differentiate it to find acceleration.
x'=1/2(1/(1+t^2))^.5*2*t
=t/((1+t^2)^.5)
Differentiate using product rule:
x"= 1/((1+t^2)^3/2)
Comments (0)
Student Expert
10th Jul, 2019
Dear VK,
x2=t2+1
taking its differentiation,
2x*(dx/dt)=2t
x*(dx/dt)=t
again taking differentiation for acceleration,
x*(d2x/dt2) + (dx/dt)2 = 1
x*(d2x/dt2) = 1 – (dx/dt)2
x*(d2x/dt2) = 1 – (t/x)2 (because x*(dx/dt)=t)
(d2x/dt2) = (1 – (t/x)2)/x
(d2x/dt2) = 1/x – t2/x3
put t2 = x2 -1
(d2x/dt2) = 1/x3
which is the required acceleration.
All the best
Hope this helps
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