Question

Derivative!

Answer 1

If a rock is thrown upward on the planet Mars with a velocity of 19 m/s, its height (in meters) after t seconds is given by H = 19t − 1.86t2.
(a) Find the velocity of the rock after two seconds.

Answer 2

My plan? Find the derivative at the point x=2. Sound good?

Answer 3

Is this good???\[19(2+h)-1.86(2+h)^2-[19(2)-1.86(2)^2]\]

Answer 4

@ganeshie8

Answer 5

the numerator looks good ^
there is also a h in the bottom right ?

Answer 6

All over h!

Answer 7

nice, go ahead carry out ur plan :)

Answer 8

Thanks man! There are other parts to this, when I complete this one, will you help guide me through the rest?

Answer 9

I'll try...

Answer 10

Also you need to put a limit ^

Answer 11

Yes! Thanks!


If a rock is thrown upward on the planet Mars with a velocity of 19 m/s, its height (in meters) after t seconds is given by H = 19t − 1.86t2.
(a) Find the velocity of the rock after two seconds.
m/s

(b) Find the velocity of the rock when t = a.
m/s

(c) When will the rock hit the surface? (Round your answer to one decimal place.)
t = s

(d) With what velocity will the rock hit the surface?
m/s

Answer 12

Why is it as h approaches 0 and not 2?

Answer 13

no wait a second

Answer 14

Why did you take out the 2+h?
How do I know to do x+h, 1+h, or 2+h?
I am self taught in this class.

Answer 15

Okay looks you're using the other definition of derivative.
Can you do me a favor ? - scratch everything above ^ lets start over...

Answer 16

Yes. I am familiar with that. Here is a ?.

In what case do you use that definition and when do you use my definition?

Answer 17

Ok!

Answer 18

Yes. Totally got that. Now the ? is (what+h)

Answer 19

Whats wrong with the one I posted above?

Answer 20

19(2+h)−1.86(2+h)2−[19(2)−1.86(2)2] all over h

Answer 21

you're mixing both the definitions

Answer 22

lets first decide on the definition to use :/

Answer 23

Let me show you what my online prof and webassign gave me to use:

Answer 24

Yes, we have two definitions. pick one

Answer 25

To the right.

Answer 26

good,

H(t) = 19t − 1.86t2.

compute : H(t+h) - H(t)

Answer 27

Sure thing:
\[19+19h-1.86(h^2+2ht+t^2)-[19t+1.86t^2]\] all over h

Answer 28

Carrying on...

Answer 29

\[\large H'(t) = \lim \limits_{h\to 0 } \dfrac{H(t+h) - H(t)}{h}\]

\[\large = \lim \limits_{h\to 0 } \dfrac{19(t+h) - 1.86(t+h)^2 - [19t-1.86t^2]}{h}\]

\[\large = \lim \limits_{h\to 0 } \dfrac{19h - 1.86(2ht) - 1.86h^2}{h}\]

\[\large = \lim \limits_{h\to 0 } \dfrac{h\left(19 - 1.86(2t) - 1.86h\right)}{h}\]

\[\large = \lim \limits_{h\to 0 }~ 19 - 1.86(2t) - 1.86h\]

\[\large = 19 - 1.86(2t)\]

Answer 30

So \(\large H'(t) = 19 - 1.86(2t)\)

Answer 31

plugin t = 2, to get the velocity after 2 seconds ^

Answer 32

Thanks! Can you show me how you just got (2ht) for your third step? I thought it was (t+h)(t+h) or t^2+2ht+h^2? Did you just omit a step?

Answer 33

My question is, I got 19+-1.86H^2-3.72HT. Where did you get the latter to cancel?

Answer 34

yes i din't show u a step !

Answer 35

\[\large = \lim \limits_{h\to 0 } \dfrac{19(t+h) - 1.86(t+h)^2 - [19t-1.86t^2]}{h}\]

\[\large = \lim \limits_{h\to 0 } \dfrac{19(t+h) - 1.86(t^2 +2ht+h^2) - [19t-1.86t^2]}{h}\]

Answer 36

Ok, so \[19t+19h-1.86t^2-3.72ht-1.86h^2-19t+1.86t^2\]

Answer 37

\[19h-3.72ht-1.86h^2\]

Answer 38

yes, you can cancel out few terms ^

Answer 39

Yes !

Answer 40

Where can I cancel 3.72ht!

(last question lol)

Answer 41

factor out \(h\) first..

Answer 42

and cancel both the h's on top and bottom

Answer 43

(19-3.72t-1.8h)

Answer 44

Yes ! take the limit by putting h = 0

Answer 45

19-3.72t

Answer 46

*pulls out hair*