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?
H = 10t − 1.86t2 differentiate wr.t to t into V= 10- (1.86*2) t
so would i use that equation when the limit of t approaches 2 of: F(a+h) -F(a)/h ?
yes, that's the long way to find derivative
yay!! we get to go to Mars today :)
okay let me try to work that out....it still confuses me to what i would plug into f(a+h) and f(a)
that is the definition of a derivitive
do you have to do this , otherwise there is a easier way Derivative of \[a^x\] is \[x a^{x-1}\]
you plug (A+h) into your function
i dont think i have to do it a specific way
so i would plug (a+h) and (a) for every t?
correct
and divide by h
okay...let me try to work out f(a+h) and f(a) and ill send you what i come up with
yes and the h haha
i'd do it for x and then replace x with a to solve for your second part but that is just me
its best to split the function into parts; since the limit of sums is the sum of the limits
okay ill try that
i think im doing some algebra incorrectly...the first term would be 10(a+H) -1.86(a+h)^2 correct? before simplifying
yes
ok...now when i take 10(a+h) ....am i distributing that 10 to both a and h? to become 10a +10h?
yes
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?
+10x+10h *
and thats for the whole numerator?
yes
ok, so all over h, and then simplify?
so it should be \[-1.86(x^2+2xh+h^2)+10x+10h-(-1.86x^2+10x)\]
all over h and then you simplify correct
some terms should cancel when you do the simplifying and at the end you should be able to factor an h out
ok...let me try to work this out
when i do the first part, the -1.86(x^2 + 2xh + h^2) ...would the second term be -3.72xh?
yes
i ended up getting: -1.86h^2 - 3.72xh +10h / h
and i still need to factor out that h
that is correct
ok..
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?
yep
just one last step
eh....what would that be haha, i know i still need to find out the actual velocity at 2 seconds, right?
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
hmm...okay
the definition of a derivative is given by this limit \[\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{Delta x}\]
where h is just delta x ,
ok
so now the last step is to get the limit as h ->0
it will cancel your h term and you are left with just x terms
so your v(t)=-3.72+10
eh...im a little lost now...sorry haha
where did i lose you haha
just this final step, im not sure what to plug into what haha
so you understand the limits h->0 to get the derivative correct
yeah
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
so in order to find the velocity at t=2 simply solve v(2)
soo...would i plug 2 into the -1.86 - 3.72x +10?
so basically your velocity function is the derivative of you position function and the derivative of your velocity function is your acceleration function
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
-3.72x+10
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
i think i get it now
\[\lim_{h \rightarrow 0}(-3.72x-1.86(0)+10)=-3.72x+10\]
so solve for when t=2 and t=a
and you'll get your answers for those two
ok, thanks! let me see what i get..
ok so the first one is 17.44
and for t=a ...
would it just be 3.72a + 10 ?
-3.72*2 = -7.44+10
oh man...good catch...been doin this for too long
t=a is correct
so, 2.56
yep =]
lastly what do you know about and object at rest and it's velocity
it would be zero, right?
yep
and for t=a, when i try to submit the answer, it still says the answer is wrong haha...just my luck
ooh you forgot the -
it'd be -3.72x+10
ooh yeah...you are right! haha so for the rock hitting the surface
would i just plug in zero?
solve h(t)=0
would i use that same equation, the -3.72x -1.86h +10 ?
no your h(t)=is the function you are given but i believe the answer is the velocity is 0
hmm...it is saying 0 in incorrect
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
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
the velocity will be zero
and the time will be 10/1.86
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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