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Use the properties of the derivative to find the following:

(a) r(t) = 9ti + (t – 1)j

(b) [u(t) – 2r(t)] = ~17i + (2t – 2)j + 2/2k

(c) [(4t)r(t)] = 72ti + (8t – 4)j

(d) -[r(t) u(t)] = 16t + 372

(e) -[r(t) x u(t)] = 24 – 23 – 1(64 – 0) + (9? – ? + 4) X

-[u(2t)] = 18i + 4j

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05:29

Use the properties of the derivative to find the following:

r(t) = 3ti + (t – 1)j, u(t) = ti + t^2j + 2t^3k

(a) r(t)3i + j

(b)[u(t) – 2r(t)]5i + (2t – 2)j + 2/k

(c)[(2t)r(t)]12ti + (4t – 2)j

(d)[r(t) u(t)]32t^2 + 7t

(e)[r(t) x u(t)][u(2t)]

07:42

Use the properties of the derivative to find the following: r(t) = ti + 4tj + tZk, u(t) = 4ti + t^2j + 3k

(a) r(t) + u(t) = i + 4j + 2tk (b) 2r(t) – u(t) = 2i + 8j – 2tj (c) 4t – 3u(t) = 12t – 3t^2j (d) -[(3t)u(t)] = -3t^3j (e) [r(t) u(t)] = 4t^2j (f) [r(t) u(t)] = 4t^2j (g) r(3t) = 3ti + 12tj + 3tk

01:37

Use the properties of the derivative to find the following: r(t) = ti + Stj + +k, u(t) = 4ti + tj + tk(a) r(t)(b) -[Zr(t) u(t)](c) d[(2t)u(t)](d) [r(t) u(t)](e) dt [r(t) u(t)]r(2t)

01:16

Use the properties of the derivative to find the following. r(t) = ti + 4tj + t^2k, u(t) = 5ti + t^2j + t^3kf) d/dt r(3t)

01:53

Find the derivative, r'(t), of the vector function r(t) = a*cos(2t)i + b*sin^4(t)j + c*cos^3(t)k.

Transcript

Here we have R of t is equal to 9Ti plus T minus 1j. U of t is equal to Ti plus T square j plus 2 by 3 T cube k. A part we have to write the derivative of this that will be 9i plus j. B part now U of t minus 2 R of t will be minus 17Ti plus T square minus 2T plus 2j minus 2 by 3 T cube k. Then d by dt of U of t minus 2 R of t will be minus 17i plus 2T minus 2 minus 2T square. For the C part 4T of R of t will be 36T square i plus 4T square minus 4T j. So then d by dt of 4T R of t will be equal to 72Ti plus 8T minus 4j d1. Now R of t dot U of t will be equal to 9T square i plus T cube minus T…

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