<p>hi
I couldn't figure out some of these problems</p>
<p>*find arc length (1/16)x^4+1/(2x^2) 2<x<5
I couldn't figure out after squaring the dy/dx and putting it in square root of
(1+d^2y/dx^2)</p>
<ul>
<li>dam is parabola h=20 w=100
and the water depth is 10ft
Find the hydro force exerted on dam</li>
</ul>
<p>also last one is the how to find the integral of</p>
<p>y=(x^3+x)^(1/3)</p>
<p>HELP WOULD BE REALLY APPRECIATED :)</p>
<p>bump bump and bumpy bump</p>
<p>for the first question it's not the second derivative it's the first derivative squared and assuming you can use a calculator, just plug it in(if not, i have no idea how you would try and calculate a function to the 6th degree by hand). </p>
<p>the second one, you can use the equation F=PA => dF=PdA. You probably should try and find a function to find the area saying it is a parabola with height 20 and width 100 is not enough since it could have more or less of a curve which affects the area. if you can find the function find where y=10 intersects the function and you can use the intersection points as bounds. P can also be represented by rho*g so i'm guessing that your answer would be (value of integral)(rho)(g).</p>
<p>third one, whew, you can factor out an x so you get :
int[((x)(x^2+1))^(1/3)]dx. from here, you can use integration by parts letting:
u=(x^2+1)^(1/3) and dv=x^(1/3)dx
so that gives:
du=[2x(x^2+1)^(-2/3)]/3dx and v=3x^(2/3)/2
since
int[udv]=uv-int[vdu]
we get
[3x^(2/3)(x^2+1)^(1/3)]/2 - x^(5/3))(x^2+1)^(-2/3) + C</p>
<p>hope that helps</p>
<p>thanks a ton man you da man</p>