Find the arc length parameter along the given curve from the point where t=0 by evaluating the integral s(t)=lv(t)ldt

Then find the length of indicated portion of the curve r(t)= 5costi+5sintj+8tk, where 0<_t<_pi

The arc length parameter along the curve, starting at t=0 is s(t)=. (Type exact answers, using radicals as needed)

The length of the indicated portion of the curve is L=___. (Type exact answers, using radicals and Pi as needed)

Find the arc length parameter along the given curve from the point where t=0 by evaluating the integral s(t)=lv(t)ldt

Then find the length of indicated portion of the curve r(t)=(2e^tcost)i+(2e^tsint)j-2e^tk, -ln4<_t<_0 s(t)=__ (Type exact answers, using radicals as needed)

The length of rage indicated portion of the curve is __ units. (Type exact answers, using radicals as needed)

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

i will with my problems and all the other things that need to   be on by the way

step-by-step explanation:

step-by-step explanation:

d would equal 12