# Arc of curve of function - Find the minimum length

This is the arc of the curve:

$$y = x^3 + x^2 - \frac{29x}{2} + 1 \\ t > 0 \\ x \in [t,t+1]$$

Find $t$ for which the length of the arc of the curve is minimum.

Should I use $\int {\sqrt{1 + [f'(x)]^2}} dx$ ?

Thank you very, but very much!

Yes, one should use the formula. Note that the integral is from $t$ to $t+1$.
Then differentiate, using the Fundamental Theorem of Calculus, and proceed as usual. After setting the derivative equal to $0$, be careful when finding square roots.
• Thank you! I tried with integral, but the integral is this: $$\int{\sqrt{1 + (3x^2 + 2x - \frac{29}{2})^2}} dx$$ – MM PP Jul 15 '16 at 6:30
• In the answer, I did not suggest evaluating the integral. A nightmare involving elliptic functions. Call the function you are integrating $H(x)$. The integral is $\int_t^{t+1} H(x)\,dx$. By the Fund. Th., the derivative of this is $H(t+1)-H(t)$. Substitute. We get $\sqrt{1+(f'(t+1))^2}-\sqrt{1+(f'(t))^2}$. Set the thing equal to $0$. So $\sqrt{1+(f'(t+1))^2}=\sqrt{1+(f'(t))^2}$. Square and simplify. We get $(f'(t+1))^2=(f'(t))^2$. Take the square roots. We get $f'(t+1)=\pm f'(t)$. Continue. – André Nicolas Jul 15 '16 at 6:46
• You are welcome. The thing is slightly unpleasant, risks of error when plugging in $t+1$, really not very nice numbers. But note that for when you take both "positive" square roots you end up with a linear equation for $t$. – André Nicolas Jul 15 '16 at 6:58
doing André Nicolas say, we have: $$f'(x)=3x^2+2x-\frac{29}{2}$$ and $$g(t)=\int_t^{t+1}{\sqrt{1 + \left(3x^2 + 2x - \frac{29}{2}\right)^2}} dx$$ $$g(t)=\int_0^{t+1}{\sqrt{1 + \left(3x^2 + 2x - \frac{29}{2}\right)^2}} dx-\int_0^{t}{\sqrt{1 + \left(3x^2 + 2x - \frac{29}{2}\right)^2}} dx$$ now we calculate $g'(t)$: $$g'(t)=\sqrt{1 + \left(3(t+1)^2 + 2(t+1) - \frac{29}{2}\right)^2}-\sqrt{1 + \left(3t^2 + 2t - \frac{29}{2}\right)^2}=0$$ $$1 + \left(3(t+1)^2 + 2(t+1) - \frac{29}{2}\right)^2-1-\left(3t^2 + 2t - \frac{29}{2}\right)^2=0$$ $$\left(3t^2+2t-\frac{29}{2}+6t+5\right)^2-\left(3t^2+2t-\frac{29}{2}\right)^2=0$$ $$2(6t+5)(3t^2+2t-\frac{29}{2})+(6t+5)^2=0$$ if $t$ distinc to $-\frac{5}{6}$, we factorizing $(6t+5)$: $$6t^2+4t-29+6t+5=0$$ finally we have: $$6t^2+10t-24=0$$ wich have the solutions: $t=-\frac{16}{12}$ and $t=3$ Finally the result is: $$t=3$$