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# instantaneous velocity

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f(x)= sqrt(4t)

Find the instantaneous velocity using f(t)- f(t0)/ t-t0

a.) t0 = 1
b.) t0= 4

Thank you!

Yura_chan  Feb 11, 2016

#1
+78577
+5

a.) t0 = 1

[f(t)- f(t0)] / [ t - t0 ] =

[√(4t)  - √ (4*1)] / [t - 1]  =

[ √(4t) - √(4)] / [ t - 1]  =

[ 2 √(t) - 2] / [t - 1]  =

2 [√(t) - 1] / [t - 1]  =

2[√(t) - 1] / [ (√(t) - 1) (√(t) + 1) ]  =

2 / (√(t) + 1)

b.) t0= 4

[√(4t)  - √ (4*4)] / [t - 4] =

[ √(4t)  - √(16)] / [ t - 4] =

[ 2 √(t) - 4 /  [t - 4]  =

2 [√(t)  - 2] / [t - 4]  =

2 [√(t)  - 2] /   [ (√(t)  - 2) (√(t)  + 2) ] =

2 / (√(t)  + 2)

CPhill  Feb 11, 2016
edited by CPhill  Feb 11, 2016
Sort:

#1
+78577
+5

a.) t0 = 1

[f(t)- f(t0)] / [ t - t0 ] =

[√(4t)  - √ (4*1)] / [t - 1]  =

[ √(4t) - √(4)] / [ t - 1]  =

[ 2 √(t) - 2] / [t - 1]  =

2 [√(t) - 1] / [t - 1]  =

2[√(t) - 1] / [ (√(t) - 1) (√(t) + 1) ]  =

2 / (√(t) + 1)

b.) t0= 4

[√(4t)  - √ (4*4)] / [t - 4] =

[ √(4t)  - √(16)] / [ t - 4] =

[ 2 √(t) - 4 /  [t - 4]  =

2 [√(t)  - 2] / [t - 4]  =

2 [√(t)  - 2] /   [ (√(t)  - 2) (√(t)  + 2) ] =

2 / (√(t)  + 2)

CPhill  Feb 11, 2016
edited by CPhill  Feb 11, 2016
#2
+209
0

In part B, how did you split the bottom x-4 into (x+2)(x-2)

I thought you could only do that if it was (x-4)^2?

Yura_chan  Feb 11, 2016
#3
+78577
+5

Note, Yura_chan......

t - 4     can be factored  as

(√(t)  - 2) (√(t)  + 2)

Because

√(t) * √(t)   - 2√(t)  + 2√(t)  (-2)*(+2)   =

t  -  4

CPhill  Feb 11, 2016
edited by CPhill  Feb 11, 2016

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