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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!

 Feb 11, 2016

Best Answer 

 #1
avatar+104712 
+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)   

 

 

 

cool cool cool                      

 Feb 11, 2016
edited by CPhill  Feb 11, 2016
 #1
avatar+104712 
+5
Best Answer

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)   

 

 

 

cool cool cool                      

CPhill Feb 11, 2016
edited by CPhill  Feb 11, 2016
 #2
avatar+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?

 Feb 11, 2016
 #3
avatar+104712 
+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

 

 

 

cool cool cool

 Feb 11, 2016
edited by CPhill  Feb 11, 2016

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