Question

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)]

Answer 1

Answer:

$i ( e^{t} - e^{-t})+ j (cost-sin t)+ k (-15t^{2})$

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

$\frac{d}{dx}(e^x) = e^x$

$\frac{d}{dx}(sinx) = cosx\\\frac{d}{dx}(cosx) = -sinx$

now differentiating with respective to 't'

$\frac{d}{dt}c_{1}+ c_{2} } = e^ t i +cost j +3t^2 k - e^-t i - sintj -18t^2 k$

taking common i, j and k terms ,we get

$\frac{d}{dt}(C_{1} +C_{2} ) = i ( e^{t} - e^{-t})+ j (cost-sin t)+ k (-15t^{2})$