Recent content by Anood

How can i find The imaginary part of (1-2i)^2-i. This is what i have done so far: (1-2i)^2 (1-2i)^-i =(1-4i+4)(1-2i)^-i

the 2nd dervitive is the following; [cos(theta)+uk*sin(theta)]^2*(w*uk*[cos(theta)-uk*cos(theta)-sin(theta)-uk*sin(theta)]/([cos(theta)+uk*sin(theta)]^2)^2

it's (n) *u(x)^(n-1) right?

what's the dervitaive of u'(x)/u(x)

i don't get the follwing line: Thus, in order to evaluate the second derivative at \theta=\theta_{k} , just keep the term including the derivative of the numerator

i see now how we get \tan(\theta_{m})=\mu_{k} what's next?

w*uk*[sin(theta)-uk*cos(theta)]/[cos(theta)+uk*sin(theta)]^2=0 i couldn't solve for theta it's very complicated. Need your help!

i'm a bit confused. Can you simplify your explanation please?

shouldn't i first take the 2nd dervitive?

is it that i have to set it equal to zero and solve for theta?

i don't think there must be a minus in front of the whole expression. what do u mean by where can the extrema of F wrt?

is what i did right? i mean squring the denomenator?

df/dtheta= w*uk*[-sin(theta)+uk*cos(theta)]/[cos(theta)+uk*sin(theta)]^2

a)In terms of theta,µk and w calculate F. b)for w=400 N and µk=.25, calculate F and theta ranging from 0 to 90 in increments of 10.Graph F versus theta. c)From the general expression in part (a) calculate the value of theta for which the value of F, required to maintain constatnt speed, is a...