|
I'm stuck on a few Calculus problems...I thought maybe some of the math whizzes at tl.net could help. Any help would be much appreciated!
1) Find the derivative of y with respect to x or t as appropriate.
y = 8xe^x - 8e^x
2) A ball is dropped from the top of a building has a height of s = 400 -162t^2 meters after t seconds. How long does it take the ball to reach the ground? What is the ball's velocity at the moment of impact?
time = ?
velocity = ?
3) Find the derivative.
p = sec q + csc q / csc q
4) Find an equation for the line tangent to the curve at the point defined by the given value of t.
x = 4 sin t, y = 4 cos t, t = 3pi/4
|
!!
totally forgot all my calc :}
1. (xe^x)' = e^x + xe^x ; figure the rest out on your own ; )
2. finding the derivative of s will give you the velocity at time t, which can be found with s = 0
3. just some manipulation of the trig functions will give you a simpler eqn i'm sure which will be easy to derive
|
#1 is just two product rules.
df * g + f * dg for the first term, and the 2nd term is really just e^x with an 8 attached to it. constants don't turn into anything. they're constant. so you have
dy/dx = 8e^x + 8xe^x -8e^x
this is a very basic and easy problem, and i'm hoping you live half way around the world from america, where the school schedule is not the same.
#3 i believe the derivative of secant sec*tan the second term simplifies to 1. so you have
dp/dq = sec(q) * tan(q)
ya, these seem very basic, so ill leave the other two to u.
|
United States1989 Posts
1) Actually it's only one real product rule, since the derivative of a constant times the function is just the derivative of the function multiplied by the constant...
y = 8xe^x - 8e^x
dy/dx = 8xe^x + 8e^x - 8e^x
Simplify on your own.
2) Basically, you're required to know that velocity is the derivative of distance; that is, ds/dt = v. From there, it's pretty easy...
s = 400 - 162t^2
0 = 400 - 162t^2
t = ?
Quadratic equation, should be easy. To determine the velocity, basically you're looking at ds/dt when s = 0, so using the time that you just figured out...
ds/dt = -324t
Again, easy.
3) I think you mean p = (sec q + csc q) / csc q...because otherwise this question would be retarded. In that case, it would simplify to...
p = (sec q / csc q) + (csc q / csc q)
p = (sec q / csc q) + 1
Do it yourself from there.
4) Basically, you're dealing with parametrics. When looking at a line, all you need is a slope and a point. How do we find the slope? The derivative of course. But we're presented with a problem, since the derivative of parametrics gives us dx/dt and dy/dt, not dy/dx. Well thanks to that genius Leibniz, we can use them as fractions...
dy/dx = (dy/dt)/(dx/dt)
So yeah, do the derivatives, plug in 3pi/4, and you have your slope. To find the point, it's even easier. Just plug in 3pi/4 into both functions and you're good to go.
I hope this helped (and also I also hope that I didn't just do your homework for you). Just ask if you have any more questions.
|
One thing I don't think has been mentioned yet is for #1), if you wanna do the derivative wrt t, you need to do chain rule, and a bunch of dx/dt's will fall out.
ie. d/dt (e^x) = e^x dx/dt
|
|
|
|
|
|
EPT
