Question

find the arc length of the curve x = t^2 + 1 y = 8t^3 + 6 on the interval -2 11 years ago

Answer 1

@freckles for this problem do i have to find dx and dy before i start setting up the length formula?

Answer 2

hey man could you help me out with this problem real quick

Answer 3

ye

Answer 4

cool

Answer 5

so do i have to get the derivative of x and y or no

Answer 6

or how do i go about solving this problem

Answer 7

hello there?

Answer 8

yes you need to find dx/dt and dy/dt

Answer 9

you know power rule and constant rule apply those

Answer 10

whats the constant rule

Answer 11

derivative of a constant is zero

Answer 12

oh yeah

Answer 13

\[\frac{d(c)}{dt}=0 \text{ where } c \text{ is constant }\]

Answer 14

so their derivative would be x = 2t and y = 24t^2 right?

Answer 15

yeah

Answer 16

\[\int\limits_{-2}^{0}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} dt\]

Answer 17

and actually you mean dx/dt=2t and dy/dt=24t^2 *

Answer 18

because x=t^2+1 and y=8t^3+6

Answer 19

yeah thats what i meant

Answer 20

thank you for clearing that up