MHB Find arc length starting from P_0

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To find the arc length function for the curve y=2x^(3/2) starting from point P_0(1,2), the derivative y' is calculated as 3√x. The integral for arc length is set up as ∫ from a to b of √(9x+1) dx, where a corresponds to the x-value of P_0, which is 1. The upper limit b remains unspecified, allowing the arc length to be expressed as a function of b. The final approach confirms the integration limits as from 1 to x.
ineedhelpnow
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find the arc length function for the curve $y=2x^{3/2}$ with starting point $P_{0}(1,2)$. how do i do this? this is what I've done so far.

$y'=3\sqrt{x}$

$1+(3\sqrt{x})^2=9x+1$

$\int_{a}^{b} \ \sqrt{9x+1},dx$

what's my a and what's my b?
 
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ineedhelpnow said:
find the arc length function for the curve $y=2x^{3/2}$ with starting point $P_{0}(1,2)$. how do i do this? this is what I've done so far.

$y'=3\sqrt{x}$

$1+(3\sqrt{x})^2=9x+1$

$\int_{a}^{b} \ \sqrt{9x+1}\,dx$

what's my a and what's my b?

Hi ineedhelpnow!

Please start new questions in new threads rather than tagging them onto existing threads. As you can see I have moved your question to a new thread.The beginning of the curve is specified as the point $P_0$. Which $x$ corresponds to it? That is your $a$.

The end point is not specified, so we can leave it as $b$, meaning we get an arc length that is a function of $b$.
 
ooookay! thanks! so i do it from 1 to x?
 
ineedhelpnow said:
ooookay! thanks! so i do it from 1 to x?

Yep.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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