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Prove:

If $$h(u,v)=f(\sin u + \cos v)$$ then $$h_u \sin v +h_v \cos u = 0.$$

Setting $f(\sin u + \cos v)= f(x)$,

$$h_u = \frac{\partial h}{\partial u} = \frac{\partial f}{\partial x}\frac{d x}{d u} = \frac{\partial f}{\partial x} \cdot \cos u,$$

$$h_v = \frac{\partial h}{\partial v} = \frac{\partial f}{\partial x}\frac{d x}{d v}= \frac{\partial f}{\partial x} \cdot (- \sin v);$$

so to prove

$$h_u \sin v +h_v \cos u = 0 $$

we observe that

$$ \left( \frac{\partial f}{\partial x} \cdot \cos u \right) \sin v + \left( \frac{\partial f}{\partial x} \cdot (- \sin v) \right) \cos u =0, $$

$$\frac{\partial f}{\partial x} \cdot 0 = 0 $$ $$\text{True}$$

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Is this right way to proceed? Is there anything to add to this? Are the notation between partial and derivative correct?

Any input is much appreciated

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    $\begingroup$ I edited your question to make the flow of the logic a little more clear. $\endgroup$ – Robert Lewis Aug 21 '17 at 3:53
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If you're trying to prove $A=B,$ then it is NOT correct to proceed as follows. $$ A = B $$ Therefore $$ C = D $$ (since $C$ is the same as $A$ and $B$ is the same as $D$). Therefore $$ E=0 $$ (for the same reason). $$ \text{True} $$ I've seen students format things like this hundreds of times. I've never heard of its being taught, yet it persists.

One way of proving $A=B$ is to write $$ A = X = Y = Z = \cdots = P = Q = R = S = B. $$ Another way is to write $$ A = B \text{ if blah blah blah blah } \ldots $$ and then try to prove blah blah blah blah.

So you could write this: $$ h_u = \frac{\partial h}{\partial u} = \frac{\partial f}{\partial x}\frac{d x}{d u} = \frac{\partial f}{\partial x} \cdot \cos u $$ $$ h_v = \frac{\partial h}{\partial v} = \frac{\partial f}{\partial x}\frac{d x}{d v}= \frac{\partial f}{\partial x} \cdot (- \sin v) $$ Therefore $$ h_u \sin v + h_v \cos u =\left( \frac{\partial f}{\partial x} \cdot \cos u\right) \sin v + \left( \frac{\partial f}{\partial x} \cdot (- \sin v) \right) \cos u =0. $$

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Close!

The procedure is essentially correct, but . . .

There are a few improvements which have been made, to wit: it's probably better, as Michael Hardy observed, not to state your desired result before it's proof without some explanatory words lest your reasoning seem circular. My other remark goes to the

notation: it would most likely be better to write

$\dfrac{df}{dx} \dfrac{\partial x}{\partial u}, \tag 1$

rather than

$\dfrac{\partial f}{\partial x} \dfrac{dx}{du}, \tag 2$

etc., since $f(x)$ is a function of the single variable $x$, which in turn depends on the two variables $u$ and $v$; the revised notation which I presented here is in accord with the customary usage of $d/dx$ versus $\partial/\partial u$ and so forth.

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    $\begingroup$ $-1$ since it is not correct. See my answer. $\endgroup$ – Michael Hardy Aug 21 '17 at 3:40
  • $\begingroup$ @MichaelHardy: a caution against MSE'ing it when too sleepy! $\endgroup$ – Robert Lewis Aug 21 '17 at 4:21

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