Whakaoti i te V=4pisqrt{x/4},s

$$V = 4 \pi \sqrt { \frac { x } { 4 } } , s$$

Whakaoti mō x

$$x=\frac{\left(\frac{V}{\pi }\right)^{2}}{4}$$

$$V\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$4\pi \sqrt{\frac{x}{4}}=V$$

Whakawehea ngā taha e rua ki te 4\pi .

$$\frac{4\pi \sqrt{\frac{1}{4}x}}{4\pi }=\frac{V}{4\pi }$$

Mā te whakawehe ki te 4\pi ka wetekia te whakareanga ki te 4\pi .

$$\sqrt{\frac{1}{4}x}=\frac{V}{4\pi }$$

Pūruatia ngā taha e rua o te whārite.

$$\frac{1}{4}x=\frac{V^{2}}{16\pi ^{2}}$$

Me whakarea ngā taha e rua ki te 4.

$$\frac{\frac{1}{4}x}{\frac{1}{4}}=\frac{V^{2}}{\frac{1}{4}\times 16\pi ^{2}}$$

Mā te whakawehe ki te \frac{1}{4} ka wetekia te whakareanga ki te \frac{1}{4}.

$$x=\frac{V^{2}}{\frac{1}{4}\times 16\pi ^{2}}$$

Whakawehe \frac{V^{2}}{16\pi ^{2}} ki te \frac{1}{4} mā te whakarea \frac{V^{2}}{16\pi ^{2}} ki te tau huripoki o \frac{1}{4}.

$$x=\frac{V^{2}}{4\pi ^{2}}$$

Whakaoti mō V (complex solution)

$$V=2\pi \sqrt{x}$$

Whakaoti mō x (complex solution)

$$x=\frac{\left(\frac{V}{\pi }\right)^{2}}{4}$$

$$|\frac{arg(V^{2})}{2}-arg(V)|<\pi \text{ or }V=0$$

Whakaoti mō V

$$V=2\pi \sqrt{x}$$

$$x\geq 0$$