2 Calc questions

1) If s(t) = 2t^3 - 21t^2 + 60t is the position function of a particle moving in a straight line, would you be able to find its total distance traveled in, say 3 seconds, by finding s(0), s(1), s(2), s(3), and calculating the absolute value between each of them and then summing those values, as opposed to differentiating the function first, setting the derivative to 0, and solving for t?

Would you get the same answer?

I've tried it and it seems to give the same answer. I was just wondering if it was true for any position function. I would say yes.

2) The function of a line is y^2 + x^3 = 9. I calculated the slope of its tangent to be -3x^2/2y. It asked us to find a point(s) so that its tangent line is y + 6x = 13. So it's slope must be -6 at that point.

I got (2,1) as a point. Are there more than one, or is that the only one?

Hello, BrownianMan!

Good question . . . but the answer is No.

Quote:

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1) If $\displaystyle s(t) \:=\: 2t^3 - 21t^2 + 60t$ is the position function of a particle moving in a straight line,
would you be able to find its total distance traveled in, say 3 seconds, by finding $\displaystyle s(0), s(1), s(2), s(3)$,
and calculating the absolute value between each of them and then summing those values,
as opposed to differentiating the function first, setting the derivative to 0, and solving for $\displaystyle t$ ?

Would you get the same answer? . . . . No

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It worked for this problem because the turning points occured at integral values of $\displaystyle t.$


Suppose we have: .$\displaystyle s(t) \;=\;4t^3 - 27t^2 + 54t$

And we want the distance traveled in the first two seconds.


By your method, we would have:

. . $\displaystyle \begin{array}{ccc}s(0) &=& 0 \\ s(1) &=& 31 \\ s(2) &=& 32 \end{array}$


Our interpretartion would be:
. . In the first second, the particle moved 31 units to the right.
. . In the next second, it moved 1 unit to the right.

Hence, the total distance moved is 32 units . . . But this is wrong!



We have: .$\displaystyle s(t) \:=\:4t^3 - 27t^2 + 54t$

. . .Then: .$\displaystyle v(t) \:=\: 12t^2 - 54t + 54$


To find turning points, solve $\displaystyle v(t) \,=\,0$

. . $\displaystyle 12t^2 - 54t + 54\:=\:0 \quad\Rightarrow\quad 6(2t^2-9t+9)\:=\:0\quad\Rightarrow\quad 6(2t-3)(t-3) \:=\:0 $

Hence, turning points occur at: .$\displaystyle t\;=\;\tfrac{3}{2},\;3$


Our table would look like this:

. . $\displaystyle \begin{array}{ccc}s(0) &=& 0 \\ s(1) &=& 31 \\ s(\frac{3}{2}) &=& 33.75\\ s(2) &=& 32 \end{array}$


In the first second, the particle moved 31 units to the right.;
In the next half-second, it moved 2.75 units to the right.
In the last half-second, it moved 1.75 units to the left.

Therefore, the total distance is: .$\displaystyle 31 + 2.75 + 1.75 \;=\;{\color{blue}35.5\text{ units}}$


Get it?

Thanks!

What about #2?

Anyone?