Time remaining:
Find the exact length of the polar curve

label Calculus
account_circle Unassigned
schedule 0 Hours
account_balance_wallet $5

r = theta ^2

0<theta<pi/4

Apr 29th, 2015

The formula is integral from 0 to pi/4 dt of sqrt(r^2(t) + (r')^2(t)).

Here r(t) = t^2, r'(t) = 2t, the integral is
int[sqrt(t^4 + 4t^2)]dt = int[t*sqrt(t^2 + 4)]dt = (1/2)int[sqrt(y + 4)]dy where y = t^2.

Further = (1/2)(2/3)[(y+4)^(3/2)] = (1/3)[(t^2 + 4)^(3/2)] with the substitution t from 0 to pi/4:

(1/3)[ ((pi/4)^2 + 4)^(3/2)  -  4^(3/2) ].

The only simplification here is 4^(3/2) = 8.

Apr 29th, 2015

Did you know? You can earn $20 for every friend you invite to Studypool!
Click here to
Refer a Friend
...
Apr 29th, 2015
...
Apr 29th, 2015
Oct 19th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer