Draw and find the Length of $x = t + 4, y = t^2 + 1$

Question

I want a confirmation for this exercise.

My try: pic

Thanks.

Answer 1

Using the formula $$l=\int_{t_0}^{t}\sqrt{\dot{x^2}+\dot {y^2}}dt=\int_{0}^t\sqrt{1+4t^2}dt$$ The indefinite integral has the form $$\frac{1}{2} \sqrt{4 t^2+1} t+\frac{1}{4} \sinh ^{-1}(2 t)$$

Answer 2

You set up the definite integral correctly: $$L=\int_4^6 \sqrt{1+y'^2}dx=\int_4^6 \sqrt{1+(2x-8)^2}dx$$ However, you are adding $2(2x-8)^2$ without reason.

Instead, denote: $2x-8=t$ to get: $$L=\int_0^4\sqrt{1+t^2}\cdot \frac12dt,$$ which you can evaluate by changing $t=\tan y$ or $t=\sinh y$ as shown here. Answer must be approximately $4.64678$ as shown here.