Write an expression involving an improper integral

nns91
Messages
301
Reaction score
1

Homework Statement


AN object moving along a curve in the xy-plane is at position (x(t),y(t)) at time t, where

dx/dt=Arcsin(1-2*e^(-t)) and dy/dt= 4t/(1+t^3)

for t>or= 0. At time t=2, the object is at the point (6,-3).

a. Let m(t) denote the slope of the line tangent to the curve at the point (x(t),y(t)). Write an expression for m(t) in terms of t and use it to evaluate lim m(t) as t approaches infinity.

b. The graph of the the curve has a horizontal asymptote y=c. Write an expression involving an improper integral that represents this value c

Homework Equations



None

The Attempt at a Solution



a. So I got m(t)= 4t/ (1+t^3)*Arcsin(1-2*e^(-t)). Then will my limit be 0 ?

b. Will the expression be: c= integral ( (dy/dt) / (dx/dt)) from 0 to infinity ??
 
Physics news on Phys.org
M(t) is (dy/dt)/(dx/dt), right? Not (dy/dt)*(dx/dt). And the limit will be 0.

To see the "end behavior" of the y values, find limit of y(t) as t -> inf.

Hint: To get y(t) you need to integrate dy/dt with respect to t, or at least write it as a function involving an integral (don't forget to find the constant of integration using the given point.)
 
Thanks. Yeah, that's what I mean, I forgot the parentheses.

For part d, will it be y= integral (dy/dt dt) from 0 to infinity ??
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top