Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: The circle can be parametrized

  1. Nov 6, 2008 #1
    For what value of b is the line y=10x tangent to the curve y=e^bx at some point in the xy-plane?
    Ans: 10/e

    Here are some more:
    Let h be the function defined by h(x)=Integration of e^(x+t) dt (limit 0 to x^2) for all real numbers. Then h'(1)=

    My ans. is coming as 2e^2-e but the correct answer available to me is 3e^2-e. Can any1 explain me how is that possible?

    let c be the circle x^2+y^2=1 oriented counterclockwise in the xy plane. What is the value of the line integral (2x-y)dx+(x+3y)dy?

    If {x} denotes the greatest integer not exceeding x, then what is the value of the integral of {x}e^-x dx (limit 0 to infinity)?

    if f is the function defined by f(x)= xe^((-x^2)-(x^-2)) [for x is not equal to 0] and
    0 [for x=0]
    at how many values of x does the graph of f have a horizontal tangent line?
  2. jcsd
  3. Nov 7, 2008 #2


    Staff: Mentor

    Re: Integration

    1. At the point of tangency (x0, 10*x0), you know that:
    • the coordinates have to satisfy y = e^(bx), and
    • the slope of the line has to equal the slope of the tangent line on y = e^(bx).
    For the first, we must have 10*x0 = e^(b*x0).
    For the second, since dy/dx = be^(bx), we must have 10 = be^(b*x0), or 10/b = e^(b*x0).

    Both of these equations have e^(b*x0) on one side, so it must be that 10*x0 = 10/b, or b = 1/x0.

    Substituting this value of b into the function equation, e^(b*x0) = 10*x0, we see that
    e^1 = 10*x0, so x0 = e/10.
    Finally, since b = 1/x0, we find that b = 10/e.

    2. There might be a way that you can do this using the Fund. Thm. of Calculus, but brute force seems to be quicker and might be easier to understand.
    You have h(x) = int(e^(x + t) dt), from t = 0 to x^2.
    So h(x) = e^x(e^(x^2) -1), carrying out the integration.
    h'(x) = e^x(2xe^(x^2)) + e^x(e^(x^2) - 1)) = e^x(2xe^(x^2) + e^(x^2)) - 1)
    h'(1) = e(2e + e - 1) = 3e^2 - e
  4. Nov 7, 2008 #3
    Re: Integration

    Thnx for the solutions!!
  5. Nov 7, 2008 #4


    Staff: Mentor

    Re: Integration

    For #3, the line integral problem, I think this is an approach (caveat: I am away from my references and it's been many years since I looked at line integrals). The circle can be parametrized as x = cos t, y = sin t, 0 <= t <= 2*pi.

    Replace x, y, dx, and dy in the integrand as above and integrate from t = 0 to t = 2pi.

    For #4, you'll need to break the integral up into a bunch of integrals, with limits of integration 0 to 1, 1 to 2, 2 to 3, and so on up to n - 1 to n. You'll eventually need to take the limit of this sum as n approaches infinity. The first integrand will be 0*e^(-x), the second will be 1* e^(-x), the third will be 2*e^(-x), and the last I showed above will be (n - 1)*e^(-x). Let S1 be the first integral, S2 the sum of the first two integrals, S3 the sum of the first three of them, and Sn the sum of the first n integrals. Take the limit of Sn as n gets large.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook