## anonymous one year ago The position of an object at time t is given by s(t) = -4 - 2t. Find the instantaneous velocity at t = 6 by finding the derivative.

1. anonymous

pleeeaaaaaase help

2. amistre64

well, what is our derivative?

3. anonymous

I have no idea how to start it that's the problem @amistre64

4. amistre64

your course materials should have a basic guideline, what have you covered?

5. amistre64

how do you define a deriative?

6. anonymous

I know but none of them look like this. They never went over a problem like this with velocity

7. amistre64

8. amistre64

does slope play into it?

9. anonymous

yes

10. amistre64

then give me your best understanding of the concept of a derivative

11. anonymous

this is what we are suppose to use but it's confusing http://www.sosmath.com/calculus/diff/der00/der00.html

12. amistre64

i agree that the technical part of it may be confusing, but the concept is pretty simple ... most concepts are. the derivative can tell us the slope of a tangent line to a curve at any given point. and it is said that a line is tangent to itself ... does this mean anything to you?

13. anonymous

Okay somewhat

14. anonymous

If i followed the formula somewhat it would be -2 but that makes no sense to me

15. amistre64

mathically, there is a long process that can be worked out with the limit of the difference quotient. $\lim_{h\to 0}\frac{f(x+h)-f(x)}{(x+h)-x}$ let f(x) be a line ... mx+b $\lim_{h\to 0}\frac{m(x+h)+b-(mx+b)}{(x+h)-x}$ $\lim_{h\to 0}\frac{mx+mh+b-mx-b}{h}$ $\lim_{h\to 0}\frac{mh+b-b}{h}$ $\lim_{h\to 0}\frac{mh}{h}$ $\lim_{h\to 0}m\frac{h}{h}$ $\lim_{h\to 0}m$ the derivative of a line, is just its slope, m

16. anonymous

f(x)=−4−2x f(x+h)=−4−2(x+h) f′(x)=limh→0f(x+h)−f(x)h→limh→0−4−2(x+h)−(−4−2x)h

17. amistre64

-2 is the derivative of your line equation. so yes. for any value of t, the derivative is a constant -2

18. anonymous

ohh geez. I overcomplicated it

19. amistre64

s(t) = -4-2t v(t) = -2

20. anonymous

Thank you!

21. amistre64

good luck :)