<p>Need your help real badly guys! Would really apreciate if you could help, thanks!</p>
<p>1) Let f(x) = 6 - x^2. For 0 < x < sqrt(6), let A(w) be the area of the triangle formed by the coordinate axes and the line tangent to the graph of f at the point (w, 6-w^2).</p>
<p>(a) Find A(1)
(b) For what values of w is A(w) a minimum?</p>
<p>2) Let f and g be functions that are differentiable for all real numbers x and that have the following properties:
(i) f '(x) = f(x) - g(x)
(ii) g '(x) = g(x) - f(x)
(iii) f(0) = 5
(iv) g(0) = 1</p>
<p>(a) Prove that f(x) + g(x) = 6 for all x.
(b) Find f(x) and g(x).</p>
<p>Really Apreciate Your Help And Thanks Again!</p>
<p>Please Help is greatly needed!</p>
<p>OK, I'll have a go.</p>
<ol>
<li> First I'll get the formula for A(w). Since f(x) = 6 - x^2, f '(x) = -2x, and in particular, f '(w) = -2w; this is the slope of the tangent line to the curve at the point (w, 6 - w^2). The equation of the tangent line, using the point-slope formula, is
y - (6 - w^2) = (-2w)(x - w)
which simplifies to y = (-2w)x + w^2 + 6. From this equation you can easily get the x and y intercepts:
y-int. = w^2 + 6
x-int. = (w^2 + 6)/(2w)
And so the area of the triangle is 1/2 * x-int * y*int which is</li>
</ol>
<p>A(w) = (1/2)(w^2 + 6)^2/(2w).</p>
<p>A(1) is 49/4.</p>
<p>To minimize the area (there's no maximum because as w --> 0 the area gets infinitely large) take the derivative of A:
A ' (w) = [16w^2(w^2 + 6) - 4(w^2 + 6)^2]/16w^2
and set it equal to zero. That happens when the numerator is zero.
I get 3w^2 - 6 = 0, or w = sqrt(2).</p>
<p>thanks a lot greennblue! Really Apreciate it. Can anyone help me with #2?</p>
<p>Problem 2.
[f(x) + g(x)]' = f '(x) + g ' (x) = f(x) - g(x) + g(x) - f(x) = 0, so
f(x) + g(x) = C. Since f(0) = 5 and g(0) = 1, C = 6. That solves (a).</p>
<p>Since f + g = 6, f(x) = 6 - g(x). Plugging into (ii) we have
g'(x) = g(x) - (6 - g(x)) = 2g(x) + 6, or
g'(x) - 2g(x) = 6. This is a first-order linear differential equation. First solve the associated homogeneous equation
g'(x) - 2g(x) = 0, by letting g(x) = ke^mx (e = base of natural log).
You find m = 2. Then you have to add a particular solution: g = 3 will work for that. Adding the homog. and particular solutions together you get
g(x) = ke^2x + 3. From the initial condition g(0) = 1 you find k = -2.
So g(x) = -2e^2x + 3, and f(x) = 6 - g(x) = 3 + 2e^2x.<br>
I went back and checked that this f and g satisfy requirements (i) - (iv).
I love math -- can you tell?</p>
<p>Dude you Rock! I sure can tell that you love math! Yo you got any good recommdation for books that will make me master BC Calculus the way you mastered Calculus?</p>
<p>Nicely worked out greennblue. Raihan, if you're looking for a challenging book, try out Stewart's calculus.</p>
<p>There are a lot of good books out there. To get really good at Calculus, you need to do two things:
1. Understand every result thoroughly, to the point where you can outline every proof. If your teacher is lazy and tends to skip over proofs, you'll have to pester him/her after class to explain them (if you can't follow the proofs in the book). Politely insist that your teacher make every step of a proof or problem clear and correct.
2. Do lots of problems, especially the hard ones. Lots more of them than the typical teacher assigns.</p>
<p>Thanks thats very good advice</p>
<p>Is there another way of doing problem #2 because I really didn't learn homogeneous equations</p>
<p>hi raihan! (do you remember me?)</p>
<p>anyway, i'll have a go at #2:</p>
<p>f'(x)= f(x)-g(x) and g'(x)=g(x)-f(x)</p>
<p>so f'(x)+g'(x)=0</p>
<p>Therefore, f(x)+g(x)= integral of [f'(x)+g'(x)]dx = k (a constant, because the derivative is 0)</p>
<p>so k=f(0)+g(0)=5+1=6</p>
<p>and f'(x)= f(x)-g(x)= f(x) + (6-(f(x))=6</p>
<p>so f(x)=6x+5
g(x)=-6x+1</p>
<p>I'm pretty sure thats right, but if anyone wants to check my work...</p>
<p>you messed up with a + sign in this line</p>
<p>
[Quote]
f'(x)= f(x)-g(x)= f(x) + (6-(f(x))=6
[/Quote]
</p>
<p>It should be:
f'(x)=f(x) - (6-f(x)) =2f(x)-6</p>
<p>write f'(x) as df/dx and solve the differential equation by integration.</p>
<p>df/dx=2f-6</p>
<p>df/(2f-6)=dx</p>
<p>Hopefully, you can take it from here. To find the value of the constant of the integration, substitute f(0)</p>