Calculus help

[Malloc-X]

can someone help me with this question plz

(a) Given the function:

S=t^3 - 6t^2 + t , where S the distant traveled by a particle at time t.

Find S and v when a =0, where v is the velocity and a is the acceleration.

(b) Given c= q^3 - 6q^2 + 12q + 18 is the total cost function of a manufacturing firm determine the firm’s marginal cost function.

[Chicokid]

:icon_arro Malloc-X

I tried working out the first problem and I'm just curious: is the equation S=t^3 - 6t^2 + t correct :icon_ques Reason is that I'm getting negative values for the velocity and the distance.

I'm thinking the equation should be S=t^3 + 6t^2 + t but I may be wrong.

[Malloc-X]

thanx for helpin me, i figured it out and its a good thing too because it came on my final exam. i did get negative answers i think when the teacher was explainin it he said something like its slowin down or changin velocity

part a i differentiated the thingy to get V, then again to get A. then i equate A to 0 and solved for T. i then use T back in hte earlier equations to find V and S.

part B, according to applied calculus by TAN all i need to do is differentiate to get marginal cost. but in the exam i did i was given the marginal cost and told to find the total cost so i had to integrate

thanx for takin ur time to help me

[Chicokid]

:)
part a i differentiated the thingy to get V, then again to get A. then i equate A to 0 and solved for T. i then use T back in hte earlier equations to find V and S.

I did the same thing...I got T=2 then used back in the S and 1st derivate which is equivalent to V and got negative answers. I know such is possible for acceleration.

Do you mind sharing how you got the answer for that question in detail :)

[Foxy Kay]

oh my God!!!!.... now i know where to go if i have a stats problem

[Malloc-X]

S=t^3 - 6t^2 + 1 oops made and error in the question there is no T at the end it should be 1

V= 3t^2 - 12t
A= 6t - 12

equate A to 0

6t - 12 = 0
6t=12
t=2
sub t in equation S

2^3 - 6(2)^2 + 1
8 - 24 + 1 = -15

sub t in V

3(2)^2 - 12(2)
12 - 24 = -12

[Foxy Kay]

wow...all i see is numbers, letters and shapes.... That is very good. I hope u pass ur exam!!!!

[Chicokid]

S=t^3 - 6t^2 + 1 oops made and error in the question there is no T at the end it should be 1

V= 3t^2 - 12t
A= 6t - 12

equate A to 0

6t - 12 = 0
6t=12
t=2
sub t in equation S

2^3 - 6(2)^2 + 1
8 - 24 + 1 = -15

sub t in V

3(2)^2 - 12(2)
12 - 24 = -12


Yeah that was my approach but I still don't understand how come the values are negative.


i did get negative answers i think when the teacher was explainin it he said something like its slowin down or changin velocity

If something is slowing down or changing velocity, it MUST be accelerating or decelarating/retarding. However the question states that acceleration is ZERO hence velocity is constant. That's why I'm lost :confused: