AP Calculus BC Problem with Solution - Arc Length - GET 800

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AP Calculus BC Problem with Solution – Arc Length

Today I will give a solution to the AP Calculus BC Arc Length problem I gave yesterday.

Level 2 Arc Length

Find the length of the arc of the curve defined by

$x(t)=\frac{1}{12}(8t+16)^{\frac{3}{2}}, \quad y(t)=\frac{t^2}{2}$

from t = 0 to t = 4.

Solution:

$\frac{dx}{dt}=(\frac{1}{12})(\frac{3}{2})(8t+16)^{\frac{1}{2}}(8)=\sqrt{8t+16}$

and

$\frac{dy}{dt}=t$

So,

$(\frac{dx}{dt})^2=8t+16$

and

$(\frac{dy}{dt})^2=t^2$

So the desired length is

$\int_0^4 \sqrt{8t+16+t^2} \ dt$

$=\int_0^4 \sqrt{t^2+8t+16} \ dt$

$=\int_0^4 \sqrt{(t+4)^2} \ dt$

$=\int_0^4 (t+4) \ dt$

$=(\frac{t^2}{2}+4t)|_0^4$

$=(\frac{4^2}{2}+4\cdot 4)$

$=8+16$

$=\bf{24}$

Notes: (1) The arc length of the differentiable curve with parametric equations x = x(t) and y = y(t) from a to b is

$\text{Arc Length}=\int_a^b \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} \ dt$

(2) We used the chain rule to compute dx/dt and a simple power rule to compute dy/dt. See problem 2 in 320 AP Calculus BC Problems for more information on the chain rule.

More AP Calculus Problems with Explanations

If you are preparing for one of the AP Calculus exams, you may want to take a look at one of the following books.

320 AP Calculus AB Problems

320 AP Calculus BC Problems

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