<p>Hi,</p>
<p>I’m reviewing for the finals and would appreciate help for the following questions. They’re quick and easy but I don’t understand them right now. It’ll be great if you provide an short explanation and not just the answer. </p>
<li> IF dy/dx = x^2 * sqr rooty, AND y(0)=1, then</li>
</ol>
<p>y= ??? (in terms of x)</p>
<li>Integral from 0 to 3 of x / square root of (x^2 + 16) with respect to dx.</li>
</ol>
<p>Thanks!</p>
<p>FOR REFERENCE, THE "[" MEANS INTEGRAL</p>
<p>Problem 1
1. dy/dx = x^2 rooty
2. distribute: dyrooty = x^2 dx
3. integrate: [dyrooty = [x^2dx
4. 2/3 y^(3/2) = 1/3 x^3
5. Plug in and solve.</p>
<p>Problem 2</p>
<p>could be trig sub</p>
<ol>
<li><p>Use a u substitution with u=x^2 + 16. So it's 1/2ln(x^2+16) I think. I didn't write it down or anything.</p></li>
<li><p>Separate the integrals or whatever it's called. So you have rt(y)dy=x^2dx and solve from there</p></li>
</ol>
<p>Thanks!</p>
<p>One more questipn:</p>
<p>The slope to the tangent of the graph of y=Arc tan (x/2) at (2, pi/4)</p>
<p>is ???</p>
<p>Isn't it y'= 1/(4+x^2)? Evaluated at whatever x you need?</p>
<p>Edit: I'm not entirely sure of that.</p>
<p>I tried doing integral #2 using a trig substitution, and it got pretty tedious. Then I realised that a better method to do it is to obtain an expansion for sqrt[x^2 + 16].</p>
<p>The tangent one:
y=arctan(x/2)
Use the chain rule to get:
y'=2/(x^2+4)
Then plug in x=2 to get y'=2/8=1/4.</p>
<p>I seem to have forgotten a 2 on the arctan one, but devious pointed it out. Number 2 isn't as hard as you guys are making it out to be.</p>
<p>sorry guys,,</p>
<p>I have one more problem which is bugging me</p>
<p>The equation of the normal lin to the curve y= cube root of (x^2-1) at the point where x=3 is ???</p>
<p>I thought it was simple but the ans was y + 2x =4. I got y= x/2 + 1/2 </p>
<p>Can anyone explain??</p>
<p>Do you mean y'=what you typed?</p>
<p>The second problem:</p>
<p>[integral(0,3)] x/sqrt.(x/x^2+16) dx</p>
<p>u = x^2+16
du/2 = xdx</p>
<p>[integral [u(0),u(3)] u^(1/2)du/2</p>
<p>1/2 [integral (16,25)] u^(1/2)</p>
<p>=</p>
<p>20.3333</p>
<p>I think your numbers are a bit off, adidasty. I had a link up to the right work, but had the mod take it away after the OP saw it.</p>
<p>Oh.. It's the integral of x/sqrt[x^2+16]. I read it as the integral of sqrt[x^2+16].</p>
<p>Yep, it isn't a trig function due to the x on top.</p>