Find the arc length of the curve on the given interval using calculus. x=arcsint y=ln(1-t2)1/2 0≤t≤1/2

Find the arc length of the curve on the given interval using calculus. x=arcsint y=ln(1-t2)1/2 0≤t≤1/2

A circle has a radius of 4. An arc in this circle has a central angle of 288°. Either enter an exacts number in terms of π or use 3.14 for π and enter your answer as a decimal...

If the radius of the circle in the figure is 20 cm and length equals 85 cm, what is the radian measure of angle?

Find the arc length of the curve c(t)=(t3/2,cos(2t),sin(2t)), 0<=t<=1.

what is the arc length of the sector watered when the sprinkler rotates through 5pi/3?

the radius is 5cm the area of the sector is 5(pi)cm^2 the centre angle is not given and the questions asks for the answer to be left in terms of pi

How long is the arc subtended by an angle of 4π/3 radians on a circle of radius 15cm?

What is the radius of a circle in which an angle of 3 radians cuts off an arc length of 30cm?

What is the radius of a circle in which an angle of 3 radians cuts off an arc length of 30 cm?

What's the arc length for the curve defined by r(t)=3sin(t)i+3cos(t)j+4tk for 0<=t<=10? (Answer: 50) r'(t)=<3cost, -3sint, 4>

Need step by step of how to do this problem

The length of an arc is 10cm. find the angle subtending by the arc if the circumference of the circle of which the arc forms part is 60cm.

I seriously need help understanding the concept

I'm totally lost here: x= tcost y= tsint -1<or= t <or=1 find the arc length... I understand the point is to find an anti-derivative using the sqrt of each term's derivative...

Find the arc length of the curve defined by r(t)=(t, sqrt(6)/2*t^2, t^3), -1<=t<=1. r'(t)=<1, sqrt(6)t, 3t^2> sqrt(1+6t^2+9t^4) But how do I simplify this?

What's the length of the curve r(t)=<2 cos t, 2 sin t, sqrt(5)> from 0<=t<=2pi?