First person to get them all right gets full points 5. Approximating numbers usi

Question

First person to get them all right gets full points

5. Approximating numbers using linearization. Ci) Find the linearization of a suitable function to approximate the number 8.99 Cii) Find the linearization of a suitable function to approximate the number 7.97 Ciii) sketch the graphs of the functions you chose and the linearizations you found in Cip and Ciip. Civ) By how much did your linearizations over- or under-estimate the numbers in Ci) and Cii)? re l is the length of the pendulum in 6. The period of a simple pendulum is given by T 27t feet, g is the constant of acceleration due to gravity, and Tis measured in seconds. Show that if AT Al the length of the pendulum changes by Al then the change in the period, AT, is given by 7. Find the critical numbers, if any, of the function: cos 23

Explanation / Answer

5)

i) Use y = sqrt (x), let x = 9, dx = -0.01, then plug in x to find y, take the derivative, plug in x and dx to find dy. The answer is y + dy.

ii) Use y = cuberoot(x), let x = -8, dx = +0.03, same process as above.

iii) I think you can graph and find the tangent line at the x values to draw :-)

iv) basically take your answer and compare it to actually plugging in sqrt(8.99) and cuberoot(-7.97). If you are answer is above you over estimated by the difference, and if below you underestimated by the difference.

6) Take the derivative to get

dT/dl = pi (l/g)^(-1/2) * 1/g = pi l^(-1/2) / g^(1/2) = pi / (lg)^1/2,

now multply by T/T to get:

pi T / (lg)^1/2 * T, and plug in the origional equation on the bottom

pi T/ [ (lg)^1/2 * 2 pi sqrt (l/g) ] = pi T / 2 l

so

dT/dl = pi T / 2l

multiply dl over and boom!

7)

f(t) = t^2 ln t

take derivative set = 0 and solve, numbers must be in domain

do product rule here:

t^2 (1/t) + ln t (2t) = 0

t + 2t ln t = 0

t(1 + 2 ln t) = 0

t = 0 (not a CN b/c not in the domain)

1 + 2 ln t = 0

ln t = -1/2

t = e^(-1/2) is the CN!

g = 2 sin x - cos 2x

g' = 2 cos x + 2 sin 2x = 0

2 cos x + 2 * 2 sin x cos x = 0 (using double angle identity here)

2 cos x (1 + 2 sin x) = 0

2 cos x = 0, so cos x = 0, so x = pi/2, 3pi/2 ...

1 + 2 sin x = 0, sin x = -1/2, so x = 7pi/6, 11pi/6...

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