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 veri

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fy the differentiation rules. d dt [c1(t) + c2(t)]

1 answer:

Zina [12.3K] · Answer 1 · 2026-03-05T08:41:32+03:00

Answer:

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

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

Step-by-step explanation:

Step 1:-

We have c1(t) = e^ t i + (sin(t))j + t³k

and c2(t) = e^−t i + (cos(t))j − 6t³k.

By adding c1(t) and c2(t):

c1(t)+c2(t) = e^ t i + (sin(t))j + t³k + e^−t i + (cos(t))j − 6t³k

Now, employing the derivative formula:

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

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

Next, differentiate with respect to 't'

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

By factoring out i, j, and k terms, we arrive at:

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