1. ## Calculus help!

it's been a really long time since my last math course (10 years) and need some help with calculus!

I have found the derivative of: 1/6 - (2000-x)/ 4(sqrt((2000-x)^2+600^2) but now I'm a bit lost.

Any help would be great!

The time required to walk to a bus stop is given by

What value of minimizes the amount of time taken? [Where represents the distance walked.]

3. ## Re: Calculus help!

Sorry, it was showing on my screen. Odd.

t=x/6 + sqrt((2000-x)^2 + 600^2)/4 is the original problem.

4. ## Re: Calculus help!

Looking of the the value of x

5. ## Re: Calculus help!

Rewriting using LaTeX, $t = \dfrac{x}{6}+\dfrac{1}{4}\sqrt{(2000-x)^2+600^2}$

Rewriting as a quadratic equation, you get $0 = 5x^2+(48t- 36,000)x+39,240,000-144t^2$

Using the quadratic formula, you get $x = \dfrac{36,000-48t \pm \sqrt{(48t-36,000)^2+20(144t^2-39,240,000)}}{10}$

Is this what you wanted?

If you wanted the minimum of t, then you can do the following:

$\dfrac{dt}{dx} = \dfrac{1}{6}+\dfrac{x-2000}{4\sqrt{(2000-x)^2+600^2}}$

To find critical points, we first find where the derivative is not defined, which occurs when the quantity in the square root is less that zero, or when the denominator of the fraction is zero. So, we want to find values for $x$ where $(2000-x)^2+600^2\le 0$. Since $(2000-x)^2+600^2 \ge 600^2$, the derivative is defined for all real $x$.

Next, find where the derivative equals zero:
$0 = \dfrac{1}{6}+\dfrac{x-2000}{4\sqrt{(2000-x)^2+600^2}}$

Solving for $x$ gives $x = 2000 \pm 30\sqrt{5}$

Checking the second derivative:

$\dfrac{d^2t}{dx^2} = \dfrac{90,000}{(x^2-4,000x+4,360,000)^{3/2}} > 0$ for all $x$, so both critical points give local minima.

Plugging in: $t(2000-30\sqrt{5}) = \dfrac{1000}{3}+\dfrac{125\sqrt{5}}{2} \approx 473.09$

$t(2000+30\sqrt{5}) = \dfrac{1000}{3}+\dfrac{145\sqrt{5}}{2} \approx 495.45$

So, the absolute maximum occurs at $x = 2000+30\sqrt{5}$.