Asked by Cecille
The position of a particle moving along the x-axis is given by s(t)=2t2+4. Use difference quotients to find the velocity v(t) and acceleration a(t), filling in the following expressions as you do so:
v(t)=limh->0 [_____ / h ] = ____
a(t)=limh->0 [ ____ / h ] = _____
ok my answer for v(t)= h(4t+h)/ h = 4t
and for a(t)= 4h/h = 4
for some reason the first part of v(t) is wrong, could someone please tell me what's wrong with it??
v(t)=limh->0 [_____ / h ] = ____
a(t)=limh->0 [ ____ / h ] = _____
ok my answer for v(t)= h(4t+h)/ h = 4t
and for a(t)= 4h/h = 4
for some reason the first part of v(t) is wrong, could someone please tell me what's wrong with it??
Answers
Answered by
Helper
v(t)= h(4t+h)/ h = 4t
This should be,
v(t)= 2h(2t + h)/h = 2(2t + h) = 4t + 2h
v(t) = (2(t + h)^2 + 4 - (2t^2 + 4))/h
v(t)=(2t^2 + 4ht + 2h^2 + 4 - 2t^2 - 4)/h
v(t) = (4ht + 2h^2)/h
v(t) = 2h(2t + h)/h
v(t) = 2(2t + h)
v(t) = 4t + 2h
Then, lim h-->0
v(t) = 4t + 2h
v(t) = 4t + 0 = 4t
Velocity, v is the derivative, f'(t).
The derivative of s(t) = 2t^2 + 4 is,
f'(t) = 4t
So, if your problem answer says v = 4t is wrong, you will have to ask your teacher.
This should be,
v(t)= 2h(2t + h)/h = 2(2t + h) = 4t + 2h
v(t) = (2(t + h)^2 + 4 - (2t^2 + 4))/h
v(t)=(2t^2 + 4ht + 2h^2 + 4 - 2t^2 - 4)/h
v(t) = (4ht + 2h^2)/h
v(t) = 2h(2t + h)/h
v(t) = 2(2t + h)
v(t) = 4t + 2h
Then, lim h-->0
v(t) = 4t + 2h
v(t) = 4t + 0 = 4t
Velocity, v is the derivative, f'(t).
The derivative of s(t) = 2t^2 + 4 is,
f'(t) = 4t
So, if your problem answer says v = 4t is wrong, you will have to ask your teacher.
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