Source
r/cheatatmathhomework•
Extremeans

Curvilinear motion question, would love some help...

I'm trying to find the velocity of a "plane in a computer game" at t=0.500s. Equation given is y=3.00+x^(-1.5). Plane starts at (1,4) and moves with constant horizontal velocity of 1.2cm/sec (Yes, cm/sec)

I tried taking the derivative WRT time, then inputting x=(1+(.5x1.2)) and Vx = 1.2cm/sec but I can't get anywhere near the right answer.

I'd really appreciate it if someone could point me in the right direction. Even just a hint at what I'm getting wrong here would be amazing.

Thanks

Posted Oct 27, 2010, 6:45 AM UTC

18
lutusp•

Okay, at first glance it seems you have a function:

y = f(x) = 3+x^-1.5

This is easy enough to differentiate. But if x is supposed to be derived from time, and if time is associated with a horizontal (x) velocity in cm/sec, then maybe it would be better to rewrite the function in terms of time, so that x = t * 1.2, like this:

y = f(t) = 3+(t * 1.2)^-1.5

Then differentiate this using the methods you've been given, at t = 0.5.

2
ExtremeansOP•

Thank you, it worked perfectly.

This is my first calculus class and I'm having difficulty with the terminology "with respect to x/t/y". It's a little hard for me to picture in my head.

*edit: Actually, I got the right answer by somehow making a mistake that I can't replicate... Arrgh. Still not able to get the right one. The derivative I got was: dy/dt=3-1.5(1.2t)^(-2.5) (1.2) For t=0.5, dy/dt=-3.45. dy/dt should equal less than 1 I think

1
niluje•

That's because lutusp made a small mistake while writing x as a function of y, yours was good: x(t)=1+1.2*t

Then you can plug that in in y(t) = 3 + (1+1.2*t)^-1.5

Now you can get dy/dt and it won't be negative.

Remark: it you want to get the "whole" velocity and not only the vertical velocity, you have to compute:

[; v(t)=\sqrt{\frac{dx}{dt}^2+\frac{dy}{dt}^2}=\sqrt{1.2^2+\frac{dy}{dt}^2} ;]
3
lutusp•

That's because lutusp made a small mistake ...

Indeed I did. I failed to notice that the x range begins at 1, not 0. Thanks!

2
ExtremeansOP•

Alright, managed to get it right with the help from both of you. My brain is mush at this point, but I think I understand it all. Thanks again!

1
lutusp•

Okay, I will explain a little further. The function being described defines the y (vertical) axis and the x (horizontal) axis, and both axes are defined by time.

The x value increases at 1.2 centimeters per second, so it's natural to express the x value in terms of time: x = 1.2 * t.

The y value is acquired with 3 + x^-1.5, but we already know that x = 1.2 * t, so we can replace x with (1.2 * t) in the function and write the final form.

Just imagine a point in two-dimensional space whose horizontal coordinate is defined by t * 1.2, and whose vertical coordinate is defined by 3 + (t * 1.2)^-1.5. This becomes easier with practice. The idea is to figure out which variable controls the process (in this cased it's time = t).

Formally, in Cartesian coordinates, the point (x,y) is defined as a function of time:

x = t * 1.2

y = 3+(t * 1.2)^-1.5

2
lutusp•

The derivative I got was: dy/dt=3-1.5(1.2t)(-2.5) (1.2)

Your result is a bit difficult to read, but unless I miss my guess, remember that certain constants drop out in derivatives, and that 3 shouldn't be there (it's not a factor of t).

Don't click this link until you're ready to see the answer.

2