# Geometrical Applications of Calculus Question

• Nov 8th 2010, 10:42 PM
Jeffrey
Geometrical Applications of Calculus Question
Could someone help me with this question?
Grant is at point A on one side of a 20m wide river and needs to get to point B on the other side 80m along the bank as shown:
________A_________________________________
||||||||||\
|||||||||| \x
|||20m|||||\
|||||||||||||\
--------------------------------B---------
|||||||||<--------80m---------->
Grants swims to any point on the other bank and then runs along the side of the river to point B. If he can swim at 7km/h and run at 11km/h, find the distance he swims (x) to minimise the time taken to reach point B.

The answer is 26m, but I'm not sure how to get that answer. Also, sorry about the picture... I tried. Thanks!
• Nov 9th 2010, 12:20 AM
Unknown008
Hm... interesting little problem if a current was also included...

Well, anyway, the time Grant takes to cover x is given by:

$\displaystyle t_1 = \dfrac{x}{7}$

The time Grant takes to cover the distance on the ground, y, is given by:

$\displaystyle t_2 = \dfrac{y}{11}$

The distance y is given by:

$\displaystyle y = 80 - \sqrt{x^2 - 20^2}$

The total time is: $\displaystyle t_1 + t_2 = T$

This all give:

$\displaystyle T = \dfrac{x}{7} + \dfrac{80 - \sqrt{x^2 - 20^2}}{11}$

$\displaystyle 77T = 11x + 560 - 7(x^2 - 400)^{\frac12}$

Differentiate with respect to x.

$\displaystyle 77\dfrac{dT}{dx} = 11 - 7(x^2 - 400)^{-\frac12} \cdot \dfrac12 \cdot 2x$

Equate this to zero:

$\displaystyle 0 = 11 - \dfrac{7x}{\sqrt{x^2 - 400}}$

$\displaystyle 11\sqrt{x^2-400}= 7x$

Square both sides, solve for x and you should get 25.92724... = 26 m