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OpenStudy (precal):

How do you find the instantaneous velocity of y at the specified value of x (given a table)?

OpenStudy (precal):

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OpenStudy (anonymous):

find the derivative of the position function at that value of x.

OpenStudy (precal):

Hard to write in the value so here it goes the x values are 0, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4 the y values are 0, 0.02, 0.08, 0.18, 0.32, 0.5, 0.72, 0.98

OpenStudy (unklerhaukus):

do you know if y(x) is a polynomial ?

OpenStudy (precal):

no, I was just given the table (no other information)

OpenStudy (henryblah):

What does x denote? Distance? Time?

OpenStudy (henryblah):

And there must be some information somewhere you might be missing. Or implicit in previous questions, examples. It might be good to give us some context.

OpenStudy (precal):

Let me type the actual instructions: Use the table to find the instantaneous velocity of y at the specified value of x. x=1 table given

OpenStudy (precal):

solution is 1 but I don't see it other than the table is approaching 1

OpenStudy (henryblah):

So it doesn't say what x and y mean? It says the instantaneous velocity of y, is y an object, is it the series of numbers? It's really unclear.

OpenStudy (precal):

It was a problem given as a semester exam review. I got confused by the instructions myself. I will have to go read up on this topic.... Thanks

hartnn (hartnn):

i think , y is the distance and x is the time take 2 points equidistant from x= 1 say x = 0.8 and x = 1.2 finding the derivative at a point is same as finding the slope of curve at that point so find the slope using these 2 points 0.8,0.32 and 1.2,0.72

hartnn (hartnn):

y is the distance and x is the time <<<<<< then dy/dx gives velocity

hartnn (hartnn):

just make sure you take equidistant points, so you can also choose this pair 0.6,0.18 and 1.4, 0.98

OpenStudy (precal):

ok thanks

OpenStudy (precal):

Got it, now I see how to do these types of problems. Thanks for the great explanation.

hartnn (hartnn):

welcome ^_^

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