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**Summary:**Trying to differentiate with respect to ## \theta ## is entangling me in cos and sec terms.

A simple problem I found, while looking for calculus practice. Roads between home and main road are 30 mph, main road is 60mph:

What is the optimum ## \theta ## to minimise journey time?

## t = \frac {distance} {speed} ##, so is a combination of the hypotenuse, and the fixed distance b minus the opposite side of the triangle formed.

In terms of ## \theta ##:

## t = \frac a {30\cos\theta} + \frac {b-atan\theta} {60} ##

## \frac {dt} {d\theta} = \frac 1 {30} asec\theta tan\theta - \frac 1 {60}{asec^2\theta} ##

Set ## \frac {dt} {d\theta} ## to zero:

$$ \frac 1 {30} asec\theta tan\theta = \frac 1 {60} asec^2\theta $$

Multiply both sides by 60:

$$ 2asec\theta tan\theta = asec^2\theta $$

Divide by a:

$$ 2sec\theta tan\theta = sec^2\theta $$

Divide by ## sec\theta ##:

$$ 2tan\theta = sec\theta $$

Alternatively:

$$ \frac {2sin\theta} {cos\theta} = \frac 1 {cos\theta} $$

Multiply by ## cos\theta ##:

$$ 2sin\theta = 1 $$

$$ sin\theta = \frac 1 2 $$

$$ \theta = sin^{-1} \frac 1 2 $$

$$ \theta = 30\deg $$

And, although I’ve been puzzling over this for over an hour, in setting out the question here, I’ve got an answer. Is it correct?

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