Question

Need help with this 2 calculus problems
I'll post the pic in a sec. C:
Please and thank you

Answer 1

Problem 7 &8

Answer 2

for avg velocity its basically slope formula using the 2 endpoints
-- velocity = change in distance / change in time

Answer 3

Can you do I) ? So I can see how its done?

Answer 4

ok interval [1,3]
\[slope = \frac{10.7 - 1.4}{3-1} = \frac{9.3}{2} = 4.65\]

Answer 5

Ohh I see. Can you do an example for problem 8? And thanks I get problem 7 now

Answer 6

ok avg velocity for interval [1,2]
\[slope = \frac{f(2) -f(1)}{2-1} = (2\sin 2\pi +3\cos 2\pi)- (2 \sin \pi +3 \cos \pi) = 3-(-3) = 6\]

Answer 7

Ohh that doesnt seem too hard. And for 7b. Is the answer 0.?

Answer 8

hmm not sure...its not zero because you can see that velocity is increasing over time
it says to use a graph, is one provided or you can plug the points into a calculator and it will estimate a graph and function for you

Answer 9

No

Answer 10

how have you been estimating instantaneous velocity in class?
im drawing a blank, sorry too used to calculus

Answer 11

They really didnt teach me. They only showed us some graphs that aren't even related to this

Answer 12

hmm ok well for #8 you can estimate it by doing a very small interval close to the point
--> [.999, 1.001] for t=1

Answer 13

but for #7, you dont have an equation to plug in the values only a limited table
you have to first estimate a relation or equation to match the table, then use that to evaluate slope for say [2.99, 3.01]

Answer 14

I got 7aa but not 7b

Answer 15

here is screenshot of using excel to make graph and make a trendline to fit data

use that equation to estimate instantaneous velocity at t=3

Answer 16

Just plug in the 3 in the equation

Answer 17

no you want velocity...change in distance/ change in time
find slope for interval [2.99, 3.01]
the smaller the interval the more accurate the instantaneous velocity

Answer 18

3.56

Answer 19

i get something closer to 6

Answer 20

How?

Answer 21

did little rounding
\[f(t) = 0.85t^{2} +.9t\]
\[f(3.01) = 10.41\]
\[f(2.99) = 10.29\]
\[slope = \frac{10.41 - 10.29}{3.01-2.99} = 6\]