# Help in Calculus

• Mar 7th 2006, 03:13 PM
frozenflames
Help in Calculus
A particle moves along the y-axis with velocity given by v(t) = t sin(t^2) for t≥ 0.

a) In which direction (up or down) is the particle moving at time t = 1.5? Why?

b) Find the acceleration of the particle at time t = 1.5. Is the velocity of the particle increasing at t = 1.5? Why or why not?

c) Given that y(t) is the position of the particle at time t and that y(0) = 3, find y(2).

d) Find the total distance traveled by the particle from t = 0 to t = 2.</html
• Mar 7th 2006, 03:41 PM
Jameson
a) What does the sign of the velocity indicate?

b)The derivative of the velocity function is the acceleration function. Look at the key word of velocity increasing. Look at the derivative of velocity for how it's changing.

c)The integral is a great way of evaluating areas of irregular shapes. The net area under the velocity function will tell you the toal distance traveled. You have not included an upper bound.
• Mar 7th 2006, 03:46 PM
frozenflames
?...........
• Mar 7th 2006, 04:55 PM
topsquark
Quote:

Originally Posted by frozenflames
A particle moves along the y-axis with velocity given by v(t) = t sin(t^2) for t≥ 0.

a) In which direction (up or down) is the particle moving at time t = 1.5? Why?

b) Find the acceleration of the particle at time t = 1.5. Is the velocity of the particle increasing at t = 1.5? Why or why not?

c) Given that y(t) is the position of the particle at time t and that y(0) = 3, find y(2).

d) Find the total distance traveled by the particle from t = 0 to t = 2.</html

a) A positive value of v(t) usually indicates the object is moving upward. A negative sign indicates it's moving downward.

b) $\displaystyle a=\frac{dv}{dt}$. So find your acceleration. If a and v have the same sign, then v is increasing. If it's a and v have opposite signs, then v is decreasing.

c) $\displaystyle y=\int v(t) \, dt$. Do the integral. Then we know that y(0) = 3, which will set the constant of integration.

d) I haven't looked very carefully, but distance travelled can be a bit tricky. You will need to see if the object has changed direction. If it has then integrate v(t) to the point in time where it turns around, then add to that the integral of v(t) for the rest of the motion. Remember, distance always accumulates, so it will never be negative.

-Dan
• Mar 7th 2006, 06:13 PM
ThePerfectHacker
Let me add something to topsquark's response.
Quote:

Originally Posted by topsquark
...the distance can be tricky to calculate...

.
The formula
$\displaystyle \int^{t_1}_{t_0}v(t)dt$ is the displacement.
$\displaystyle \int^{t_1}_{t_0}|v(t)|dt$ is the distance.
• Mar 8th 2006, 04:17 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
Let me add something to topsquark's response. .
The formula
$\displaystyle \int^{t_1}_{t_0}v(t)dt$ is the displacement.
$\displaystyle \int^{t_1}_{t_0}|v(t)|dt$ is the distance.

10 years teaching Physics and I never thought of that! You learn something new every day. :)

-Dan
• Mar 8th 2006, 04:35 AM
Jameson
Yes. Finding distance is tricky. To the OP, consider the sine wave. Just the general one. Let's say that the velocity of a particle can be represented by y=sin(x). Now you want to find the distance covered from 0 to $\displaystyle 2\pi$.

We know that displacement is: $\displaystyle \vec{s}=\int_{t}vdt$

but as the posters above have shown, distance is $\displaystyle d=\int_{t}|vdt|$. What does this mean?

Well look at the area of the sine function from 0 to $\displaystyle \pi$. It's positive. So integrating over this region will show a positive displacement and distance. But from $\displaystyle {\pi,2\pi}$ the area of the function is negative. This means that the displacement will be negative, but in order to get the distance over that region, treat the area as positive and to your final answer for this example.

$\displaystyle D=\int_{0}^{\pi}sin(x)dx+|\int_{\pi}^{2\pi}sin(x)d x|$
• Mar 8th 2006, 04:30 PM
Rich B.
Greetings:

Another way of viewing the "negative area" business is as the area bounded by the graphs of two functions. We have a theorem that asserts: "If fand g are functions of a single variable,x, such that f > g for all x on some interval [a,b], then the area bounded by the respective graphs is given by the definite integral, int(f)dx - int(g)dx = int (f-g)dx, from a to b.". Noting that the x-axis determines the function f(x)=0, where f(x) > g(x) for any g(x) residing quadrant(s) three and/or four, the bounded area is readily calculated by int(f)-int(g) = int(f-g) = int(0-g) = -int(g)dx.

For example, the area bounded by g(x)=ln(x) and f(x)=0 on [1/e,e] is given by
int(0-ln(x)) = -int(ln(x));[1/e,1] + int(ln(x));[1,e] = -x[ln(x) - 1]; [1/e,1] + x[ln(x) - 1]; [1, e], which simplifies to 2[1-1/e] = 1.264 units^2 to three places.

Earth shattering? Not even close. But there is something to be said for generalization.

Regards,

Ricn B.