<p>i would appreciate any help</p>
<p>You could always just post your questions. This is a forum for discussion...</p>
<p>this is my question</p>
<p>f(x) = ksin(kx), where k is a positive constant</p>
<p>a) find area of region bounded by one arch of the graph of f and the X axis</p>
<p>b)find the area of the triangle formed by the X axis and the tangents to one arch of f at the points where the graph of f crosses the x axis</p>
<p>(a) Integration of f(x) = ksin(kx) , k>0 is pretty simple. Just make sure you restrict your limits of integration and you'll get the right answer.</p>
<p>(b) Find the slope of the tangents, and write them out in the form y=mx+b, find where they intersect. The height of that triangle is the y value of the point, and the width of the triangle is half the period of f(x) = ksin(kx).</p>
<p>Maybe they arne't the easiest solutions, but they'll work.</p>
<p>Draw pictures, they'll help.</p>
<p>i get part A but im lost on part B, can you do it and show your work please</p>
<p>Sorry, but I'm not going to do the problem for you, but I will try to help explain it to you.</p>
<p>The function f(x) = ksin(kx) is just a stretched out sin wave. By "one arch" of f, they mean one section with f above the x-axis. You'll notice that it is an arch. They want you to consider the two tangents to that arch where it intersects the x-axis (think derivative at those points for slope). </p>
<p>If you draw those two tangents, they will intersect above the arch and form a triangle. The base of the triangle will be the width of that arch on the x-axis.</p>
<p>Again, draw a picture, it will help. Good luck.</p>
<p>for part A would the answer be in terms of x or would it be a numerical answer, when integrating the boundaries are from 0 to x, right?</p>
<p>Since k is a constant, each arch of f(x) is going to have the same area. So it doesn't matter which one you pick.</p>
<p>Since sin(0)=0, the first arch starts at x=0 and goes until f(x)=ksin(kx) is equal to 0 again. This is actually half the period of the function. So all you have to do is find out what that value of x is, and then integrate from 0 to period/2 (i.e. the next value of x that makes f(x)=0).</p>
<p>The answers for both parts (a) and (b) will be in terms of k and pi only.</p>
<p>for part A i get:
-cos(kpi) - 1</p>
<p>is that right?</p>
<p>for part B i get:
pi^2 / 4</p>
<p>could someone tell me if i got the right answers</p>
<p>I'll check it in a bit.</p>
<p>okay thanks</p>
<p>(a) You're close. Remember that d/dx ( cos(x) ) = -sin(x) , so don't forget that negative sign. </p>
<p>Also, your period is wrong. Let T be the new period. Consider g(x) = sin(x):
Then g(2pi) = sin(2pi) = 0
So we wan't f(T) = g(2pi) = 0, i.e. ksin(kT) = sin(2pi) = 0
i.e. kT = 2pi, so T = 2pi/k.</p>
<p>Or you just forgot to cancel that k inside of your cos(kpi) term (see below).</p>
<p>Remember now that the positive arch of f only spans half of that period (for the other half it's negative). So we integrate from 0 to T/2.</p>
<p>cos(k(T/2)) = cos(k*pi/k) = cos(pi) = -1</p>
<p>That should get you the answer for part (a), which should be 2.</p>
<p>(b) Perfect. Nice work.</p>
<p>Good luck on part (a), you're almost there.</p>
<p>i got it, thanks, i appreciate your help :)</p>