Let c1(t) = eti + (sin(t))j + t3k and c2(t) = e−ti + (cos(t))j − 6t3k. Find the stated derivatives in two different ways to verify the differentiation rules. d dt [c1(t) + c2(t)]

Step-by-step explanation:

Step 1:-

Given c1(t) = e^ t i + (sin(t))j + t^3k

and c2(t) = e^−t i + (cos(t))j − 6t^3k.

adding c1(t)+c2(t) = e^ t i + (sin(t))j + t^3k+ e^−t i + (cos(t))j − 6t^3k

by using derivative formula

now differentiating with respective to 't'

taking common i, j and k terms ,we get

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

i dont know

step-by-step explanation: