A skate ramp is shaped like the parabola $y = x^2$, between $x = 0$ and $x = 1$, $[0,1]$ - seen from the side.
A) Let $L$ be the length of the graph of the function $f(x) = x^2$, and show that $$ L = 1/2 * \int_0^2 \sqrt{1+t^2}dt $$
B) The model of the skate ramp shows the length to be 2 meters, and the width to be 1 meter. The profile of the ramp follows the function $f(x) = x^2$ for $x$ in $[0,1]$. What is the area of the ramp, expressed with $L$?
Here's a picture of the model of the skate ramp:
Now, I'm familiar with the arc length formula, i believe.. $t^2$ here I guess is $f'(x)^2$, so $2x$? But why is the integral from 0 to 2, and not from 0 to 1? and where does the 1/2 part come from?
My first thought is that 1/2 multiplied with the integral from $0$ to $2$ is the same as just the integral from $0$ to $1$. Is that correct?
And for B) Should I then do the integral to get the area? My question is then, won't that area just be the area on the side, on not the ramp (the part u can skate on) itself?
$L_{rmp}$ is length of the ramp $$ L_{rmp} = \int dl$$ $$ dl = \sqrt{dx^2 + dy^2}$$ $$ L_{rmp} = \int \sqrt{dx^2 + dy^2} = \int \sqrt{1 + (d_xy)^2} dx $$ $$ y = x^2 \Rightarrow d_x y = 2x$$ $$ L_{rmp} = \int \sqrt{1 + 4 x^2} dx$$ As we knows $x \in [0, 1]$ $$ L_{rmp} = \int_0^1 \sqrt{1 + 4 x^2} dx \ \ (1)$$ $$t = 2x \Rightarrow dt = 2 dx \Rightarrow dx = \cfrac{dt}{2}$$ $$ x_0 = 0, x_1=1 \Rightarrow t_0 = 0, t_1 = 2$$ $$ L_{rmp} = \cfrac{1}{2} \int_0^2 \sqrt{1 + t^2} dt \approx 1.48$$ As width of the ramp is a const $$S_{rmp} = width * L_{rmp} = 2 * L_{rmp} = \int_0^2 \sqrt{1 + t^2} dt = 2.96$$