Calculus
Graph the curve and find its exact length. x = e^t + e^-t, y = 5 - 2t, from 0 to 3 Length = Integral from 0 to 3 of: Sqrt[(dx/dt)^2 + (dy/dt)^2] dx/dt = e^t - e^-t, correct? dy/dt = -t^2 - 5t, correct? So: Integral from 0 to 3 of Sqrt[(e^t - e^-t)^2 + (-t^2 - 5t)^2] Then what ...
calc: simpson's rule & arc length
i'm still getting this question wrong. please check for my errors: Use Simpson's Rule with n = 10 to estimate the arc length of the curve. y = tan x, 0 <or= x <or= pi/4 .. this is what i did: y' = sec(x)^2 (y')^2 = [sec(x)^2]^2 [f'(x)]^2 = sec(x)^4 In...
calc: arc length
Posted by COFFEE on Monday, June 11, 2007 at 11:48pm. find the exact length of this curve: y = ( x^3/6 ) + ( 1/2x ) 1/2 <or= x <or= 1 im looking over my notes, but i'm getting stuck. here's my work so far: A ( 1 , 2/3 ) B ( 1/2 , 49/48 ) y' = [1/6 (3x^2)] + [...
calc: arc length
find the exact length of this curve: y = ( x^3/6 ) + ( 1/2x ) 1/2 <or= x <or= 1 im looking over my notes, but i'm getting stuck. here's my work so far: A ( 1 , 2/3 ) B ( 1/2 , 49/48 ) y' = [1/6 (3x^2)] + [1/2 (-1x^-2)] y' = ( x^2 / 2 ) - ( x^-2 / 2 ) (y&#...
Math/Euler's Method
Consider a cooling cup of coffee whose initial temperature is 205°. The room temperature is held at 70°. Suppose k = 1/16. Let y be the temperature, and y' its time derivative. ----------------------------------- I have the differential equation: y' = (-1/16)(y...
Math/Calculus
How would I integrate the following: (2x^2 + 5)/((x^2+1)(x^2+4))dx I think I would start with making it a sum of two partial fractions.
Math/Calculus
How would I integrate the following by parts: Integral of: (x^2)(sin (ax))dx, where a is any constant. Just like you did x^2 exp(x) below. Also partial integration is not the easiest way to do this integral. You can also use this method. Evaluate first: integral of sin(ax)dx =...
Math/Calculus
How would I evaluate the following integral by using integration by parts? Integral of: (t^3)(e^x)? You mean (x^3)(e^x)? x^3 exp(x) dx = x^3 d[exp(x)] = d[x^3 exp(x)] - exp(x) d[x^3] = d[x^3 exp(x)] - 3 x^2 exp(x) dx So, if you integrate this you get x^3 exp(x) - 3 Integral of...
Math/Calculus
How would I solve the following integral with the substitution rule? Integral of: [(x^3)*(1-x^4)^5]dx Put 1-x^4 = y Then -4x^3 dx = dy Integral is then becomes: Integral of -1/4 y^5 dy ok, thanks a lot! I got it now.
Math/Calculus #2
Integrate: 1/(x-sqrt(x+2) dx I came up with: (2/3)(2*ln((sqrt(x+2))-2)+ln((sqrt(x+2))-1)) but it keeps coming back the wrong answer even though I integrated correctly. Is there a way to simplify this answer, and if so, how? I found: Ln[x-sqrt(x+2)] + 1/3Ln[(sqrt(x+2)-2)/(sqrt(...
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