Applications of Differential Equations - Projectile Motion - Nedd help

[squance]

Need help.I can seem to do applications questions for second order DE's but I can cope with first order.

The height of a projectile fired vertically up from the ground with an initial speed of 50m/s satisfies the differential equation



Find the height at any time t.

(for that part I got a an equation that was like but I think its wrong....)　The answer by the way for that part is
What is the maximum height reached?
And for this part its (but I can't get this answer until I know how to do the first part!!!)

[enwiabe]

"fired vertically up from the ground with an initial speed of 50 m/s" are your initial conditions. That's essentially telling you that at t = 0, h = 0 and also at t = 0, velocity = 50. Velocity is the derivative of distance with respect to time, or in this case, height with respect to time. So, when t = 0, velocity = dh/dt = 50. You then sub in those points after you integrate to dh/dt and h to get the 'c' and 'd' respectively.

As for the maximum height, you'll notice that it's an inverted parabola. You have two options. You can solve the derivative of h with respect to time (dh/dt) = 0 to get the maximum and then sub in the value of 't' you get into your h equation, or you can use symmetry and note that the maximum will always be the mid-point of the two axial t-intercepts and get your 't' value from there and sub in.

Try that If you're still in trouble, someone who knows latex will be along soon to provide a worked solution

[squance]

Thankyou enwiabe!!!
I managed to solve it!!!!
(it was easy...I can't believe Im so stupid...I just have issues with word problems in general...)

Oh yeah. I have another problem but it is sooo from long ago and I haven't been able to solve it. Its related rates problem.

Find the maximum possible area of a rectangle which has its base on the x -axis and its upper vertices on the semi-circle ,  with a domain from -3 to 3.

Modified by enwiabe: Appended a /tex tag.

[enwiabe]

Squance, that's not a semi-circle. That's a square root function

[squance]

Squance, that's not a semi-circle. That's a square root function

Whoopsies daisies!! I just fixed it up now (i was just copying and pasting from the latex webiste and I forgot to change the figures... I fixed it up now)

[enwiabe]

Okely. So if you visualise this you have a circle in the x-y plane, with lines shooting up from the x-axis to intersect the circle. That forms a section of a circle. The rectangle is then made by joining the y-intercepts.

That was really vague. Picture the rectangle, with lines extending up from x = -1 and x = 1 to where they intersect the circle at y = sqrt(9 - (+/-1)^2) = sqrt(. If you can see that we're already halfway there. Anyway, the area of that rectangle at those values of x and y is base * height = |2*1(base)*8(height)| = |2x*y| (there are magnitude signs because 'x' can be plus or minus and area is always positive)

Now, let's not give x a numeric value. Let's just keep x = x and y = sqrt(9 - x^2). Because to get the area, it's just going to be 2*x*y or 2*x*sqrt(9-x^2).

Thus, A = 2Xsqrt(9-X^2)
We want to maximise the area, so we need to differentiate that. Using product rule:

dA/dx = 2sqrt(9 - X^2) + (1/2)*2X((9 - X^2)^(-1/2))*-2X
         = 2sqrt(9 - X^2) - 2x^2/((9 - X^2)^(1/2))

Area is maximised when dA/dx = 0
So, 2sqrt(9 - X^2) - 2x^2/((9 - X^2)^(1/2)) = 0
                                         2sqrt(9 - X^2) = 2X^2/((9 - X^2)^(1/2))
                                                  9 - X^2 = X^2
                                                      2X^2 = 9
                                                        X^2 = 9/2
                                                           X  = (+/-)3/sqrt(2)

You can take either value because the area will be the magnitude. But, for easiness, let's take X = 3/sqrt(2) remember that Area = |2*x*y| = |2*x*sqrt(9 - X^2) = 2*(3/sqrt(2))*sqrt(9 - (3/sqrt(2))^2)
         = 3*sqrt(2) * sqrt(9 - 9/2)
         = 3*sqrt(2)*sqrt(9/2)
         = 3*sqrt(2*9/2)
         = 3*sqrt(9)
         = 3*3
         = 9 square units.

Sorry for no latex! Feel free to ask any questions about this, it probably looks real confusing

[Mao]

enwiabe learn latex

[squance]

Thanks again enwiabe!!

I now have another question...its supposed to be simple but I keep getting different answers

Find the general solution to this differential equation:



I got two different answers

one of them was y= sin(loge |x| + c) when I did the dy/dx way and the other one was y=sin(loge(x/4)) when i did it the dx/dy way...

Can someone help me please?

[dcc]

One approach:





Integrating both sides leaves:



Interestingly enough, it could be possible that both of your answers are correct.

Remember, while the following may seem different:





It is important to remember that:

, and ln(4) is a constant! so your answers may both be correct

[squance]

There are two different answers??? NOOOOO!!!

I like it better if there was just one answer.

[dcc]

Although it does ask for the general solution, so perhaps the answer i gave (minus the c) would be the correct one.