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The following differentiated implicitly with respect to $\theta$:

$3x = \tan \, \theta $

The book says

$3 dx = \sec^2 \theta \, d \theta$

One could start the calculation like this (I think):

$ \frac{d}{d \theta}3x = \frac{d}{d \theta} \tan \, \theta $

From there I'm not really sure about the steps.

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    $\begingroup$ The book is wrong, the answer is 3 dx = sec(theta)^2 dtheta. $\endgroup$ – quanta Apr 19 '11 at 11:02
  • $\begingroup$ Sorry, typo. I'll fix. $\endgroup$ – Algific Apr 19 '11 at 11:07
  • $\begingroup$ you can do any of the ways, it gives you the same result. $\endgroup$ – amul28 Apr 19 '11 at 11:14
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I don't know what you are asking but I will explain how I view this sort of thing.

What is meant by $$3x = \tan(\theta)$$ is that both sides are the same function of $x$ or $\theta$ and $x$ and $\theta$ are related somehow. Writing out the relation explicitly we have $$3x = \tan(\theta(x)).$$ Now define $f(x) = 3x$ and $g(x) = \tan(\theta(x))$, this equation means that $f = g$.

Applying the derivative operator to both sides $f' = g'$ we have $$3 = \theta'(x)\sec(\theta(x))^2$$ (by the chain rule and derivative of tan = sec^2). Now you can write it as a differential $$3 \mathrm{d}x = \sec(\theta(x))^2 \theta'(x) \mathrm{d}x = \sec(\theta)^2 \mathrm{d}\theta$$ since $\theta'(x) \mathrm{d}x = \mathrm{d}\theta$.

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    $\begingroup$ I just leveled up! Thank you! $\endgroup$ – Algific Apr 19 '11 at 14:54
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This is differential notation. If $y=f(x)$, then it is customary to write $dy=f'(x)dx$. What this says is that a small change in $x$, $dx$, produces an approximate change on $y$ of $f'(x)dx$. In your instance you have $x=f(\theta)$ and so $dx=f'(\theta)d\theta$. The concept comes into play when you talk about linear approximation and you are using the differentials to approximate error. So you have

$$ x=\frac{1}{3}\tan\theta $$

which gives

$$ dx=\frac{1}{3}\sec^2(\theta) d\theta. $$

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It is simply that $$\frac{d}{d \theta} \left( 3x \right) = \frac{d}{d \theta} \left( \tan \theta \right)$$

Then we know that $$\frac{d}{d \theta } \left( \tan \theta \right) = \mbox{sec}^2 \theta $$

and

$$\frac{d}{d \theta} \left( 3x \right) = 3 \frac{dx}{d \theta}$$

(Can you see why?)

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  • $\begingroup$ Why is that last part equal to 3 dx? $\endgroup$ – Algific Apr 19 '11 at 11:31
  • $\begingroup$ @Algific If you have a look at quanta's answer - I could really write $3 x(\theta)$. Then it should be clear $\endgroup$ – Juan S Apr 19 '11 at 11:33

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