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Mathematics 63 Online
OpenStudy (anonymous):

If a rock is thrown upward on the planet Mars with a velocity of 10 m/s, its height (in meters) after t seconds is given by H = 10t − 1.86t2. Find Velocity after 2 seconds. Find velocity of rock when x=a. When rock hits surface?

OpenStudy (anonymous):

H = 10t − 1.86t2 differentiate wr.t to t into V= 10- (1.86*2) t

OpenStudy (anonymous):

so would i use that equation when the limit of t approaches 2 of: F(a+h) -F(a)/h ?

OpenStudy (anonymous):

yes, that's the long way to find derivative

OpenStudy (amistre64):

yay!! we get to go to Mars today :)

OpenStudy (anonymous):

okay let me try to work that out....it still confuses me to what i would plug into f(a+h) and f(a)

OpenStudy (anonymous):

that is the definition of a derivitive

OpenStudy (anonymous):

do you have to do this , otherwise there is a easier way Derivative of \[a^x\] is \[x a^{x-1}\]

OpenStudy (anonymous):

you plug (A+h) into your function

OpenStudy (anonymous):

i dont think i have to do it a specific way

OpenStudy (anonymous):

so i would plug (a+h) and (a) for every t?

OpenStudy (anonymous):

correct

OpenStudy (amistre64):

and divide by h

OpenStudy (anonymous):

okay...let me try to work out f(a+h) and f(a) and ill send you what i come up with

OpenStudy (anonymous):

yes and the h haha

OpenStudy (anonymous):

i'd do it for x and then replace x with a to solve for your second part but that is just me

OpenStudy (amistre64):

its best to split the function into parts; since the limit of sums is the sum of the limits

OpenStudy (anonymous):

okay ill try that

OpenStudy (anonymous):

i think im doing some algebra incorrectly...the first term would be 10(a+H) -1.86(a+h)^2 correct? before simplifying

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

ok...now when i take 10(a+h) ....am i distributing that 10 to both a and h? to become 10a +10h?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

okay...so i have: 10a +10h -1.86(a+h)^2 ...which is -1.86(a^2 +2ah +h^2)...? so i would i just distribute that -1.86 to each a^2, 2ah, h^2?

OpenStudy (anonymous):

+10x+10h *

OpenStudy (anonymous):

and thats for the whole numerator?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

ok, so all over h, and then simplify?

OpenStudy (anonymous):

so it should be \[-1.86(x^2+2xh+h^2)+10x+10h-(-1.86x^2+10x)\]

OpenStudy (anonymous):

all over h and then you simplify correct

OpenStudy (anonymous):

some terms should cancel when you do the simplifying and at the end you should be able to factor an h out

OpenStudy (anonymous):

ok...let me try to work this out

OpenStudy (anonymous):

when i do the first part, the -1.86(x^2 + 2xh + h^2) ...would the second term be -3.72xh?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

i ended up getting: -1.86h^2 - 3.72xh +10h / h

OpenStudy (anonymous):

and i still need to factor out that h

OpenStudy (anonymous):

that is correct

OpenStudy (anonymous):

ok..

OpenStudy (anonymous):

so.... h(-1.86h - 3.72x +10) / h ...where the h's would cancel....so you are just left with what is in the parenthesis?

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

just one last step

OpenStudy (anonymous):

eh....what would that be haha, i know i still need to find out the actual velocity at 2 seconds, right?

OpenStudy (anonymous):

yes but that isn't your derivative you remember the point of doing this was that when using the definition of a derivative you are taking the limit as h->0 so that is why you must try to cancelt the h on the bottom

OpenStudy (anonymous):

hmm...okay

OpenStudy (anonymous):

the definition of a derivative is given by this limit \[\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{Delta x}\]

OpenStudy (anonymous):

where h is just delta x ,

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

so now the last step is to get the limit as h ->0

OpenStudy (anonymous):

it will cancel your h term and you are left with just x terms

OpenStudy (anonymous):

so your v(t)=-3.72+10

OpenStudy (anonymous):

eh...im a little lost now...sorry haha

OpenStudy (anonymous):

where did i lose you haha

OpenStudy (anonymous):

just this final step, im not sure what to plug into what haha

OpenStudy (anonymous):

so you understand the limits h->0 to get the derivative correct

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

alright so if your teacher hasn't explained to you the h(t) is your position function as it will tell you where the ball is located if you take the derivative of h(t) *which you did*, you get the velocity function which will tell you the direction and speed it is going lastly (you do not have to do this) if you take the derivative of what you have you get your acceleration function

OpenStudy (anonymous):

so in order to find the velocity at t=2 simply solve v(2)

OpenStudy (anonymous):

soo...would i plug 2 into the -1.86 - 3.72x +10?

OpenStudy (anonymous):

so basically your velocity function is the derivative of you position function and the derivative of your velocity function is your acceleration function

OpenStudy (anonymous):

when you put 0 into h you should get -1.86(0) so that should have cancelled out you put 2 into -3.72+10

OpenStudy (anonymous):

-3.72x+10

OpenStudy (anonymous):

ok....ahh...on my paper i was doing my work on i left the "h" off of the 1.86....thats why i wasnt seeing it haha

OpenStudy (anonymous):

i think i get it now

OpenStudy (anonymous):

\[\lim_{h \rightarrow 0}(-3.72x-1.86(0)+10)=-3.72x+10\]

OpenStudy (anonymous):

so solve for when t=2 and t=a

OpenStudy (anonymous):

and you'll get your answers for those two

OpenStudy (anonymous):

ok, thanks! let me see what i get..

OpenStudy (anonymous):

ok so the first one is 17.44

OpenStudy (anonymous):

and for t=a ...

OpenStudy (anonymous):

would it just be 3.72a + 10 ?

OpenStudy (anonymous):

-3.72*2 = -7.44+10

OpenStudy (anonymous):

oh man...good catch...been doin this for too long

OpenStudy (anonymous):

t=a is correct

OpenStudy (anonymous):

so, 2.56

OpenStudy (anonymous):

yep =]

OpenStudy (anonymous):

lastly what do you know about and object at rest and it's velocity

OpenStudy (anonymous):

it would be zero, right?

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

and for t=a, when i try to submit the answer, it still says the answer is wrong haha...just my luck

OpenStudy (anonymous):

ooh you forgot the -

OpenStudy (anonymous):

it'd be -3.72x+10

OpenStudy (anonymous):

ooh yeah...you are right! haha so for the rock hitting the surface

OpenStudy (anonymous):

would i just plug in zero?

OpenStudy (anonymous):

solve h(t)=0

OpenStudy (anonymous):

would i use that same equation, the -3.72x -1.86h +10 ?

OpenStudy (anonymous):

no your h(t)=is the function you are given but i believe the answer is the velocity is 0

OpenStudy (anonymous):

hmm...it is saying 0 in incorrect

OpenStudy (anonymous):

i may have worded the question odd, there are two more things the problems asks: When will the rock hit the surface? and At what velocity will the rock hit the surface

OpenStudy (anonymous):

because to check you can make your original fucntion 0 which will give you 2 times t=0 and t=-10/1.86 now at t=0... that is when you initially are throwing the ball up so t=-10/-1.86 is when it hits the ground ... if you put that back into the equation you get v(-10/-1.86 =-10+10 = 0

OpenStudy (anonymous):

the velocity will be zero

OpenStudy (anonymous):

and the time will be 10/1.86

OpenStudy (anonymous):

hmm. yeah that makes sense to me but apparently it doesnt make as much sense to this online hw program haha, thank you for your help on this problem, i appreciate it!

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