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What is the instantaneous rate of change for #f(t)= sqrt (5t+1)# when t=1?

Answer 1

Ezra Adams

#(5sqrt6)/12#

The #color(blue)"instantaneous rate of change"# is the value of f'(t) at t = 1.

differentiate f(t) using the #color(blue)"chain rule and power rule"#

#f(t)=sqrt(5t+1)=(5t+1)^(1/2)#

#rArrf'(t)=1/2(5t+1)^(-1/2).d/dt(5t+1)#

#=5/2(5t+1)^(-1/2)=5/(2(5t+1)^(1/2))=5/(2sqrt(5t+1))#

#rArrf'(1)=5/(2sqrt6)=5/(2sqrt6)xxsqrt6/sqrt6=(5sqrt6)/12#

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Answer 2

Emily Ammerman

To find the instantaneous rate of change for ( f(t) = \sqrt{5t + 1} ) when ( t = 1 ), we need to find the derivative of ( f(t) ) with respect to ( t ) and then evaluate it at ( t = 1 ).

First, we find the derivative:

[ f'(t) = \frac{d}{dt}\left(\sqrt{5t + 1}\right) = \frac{1}{2\sqrt{5t + 1}} \cdot \frac{d}{dt}(5t + 1) = \frac{5}{2\sqrt{5t + 1}} ]

Now, we evaluate the derivative at ( t = 1 ):

[ f'(1) = \frac{5}{2\sqrt{5(1) + 1}} = \frac{5}{2\sqrt{6}} = \frac{5\sqrt{6}}{12} ]

Therefore, the instantaneous rate of change for ( f(t) = \sqrt{5t + 1} ) when ( t = 1 ) is ( \frac{5\sqrt{6}}{12} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.