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Question

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ganeshie8 · Answer

find the total differential dP

ganeshie8 · Answer

$$\Delta P =   P_V  \Delta V + P_R \Delta R$$

ganeshie8 · Answer

$P_V$ and  $P_R$ are partial derivatives

ganeshie8 · Answer

find them plugin above

ganeshie8 · Answer

you don't need second partial derivatives

ganeshie8 · Answer

$$P = \dfrac{V^2}{R}$$

$$P_V = \dfrac{2V}{R}$$

$$P_R = \dfrac{-V^2}{R^2}$$


yes ?

ganeshie8 · Answer

evaluate them at V=217, R = 19

ganeshie8 · Answer

we are doing that only

ganeshie8 · Answer

$$\Delta P =   P_V  \Delta V + P_R \Delta R$$

$$\Delta P =   \frac{2\times 217}{19} (-2.25) + \frac{-217^2}{19^2}(-0.04)$$

simplify

ganeshie8 · Answer

Oh I see what you mean

ganeshie8 · Answer

Okay lets fix the mistake, below gives you $dP$ not $\Delta P$

$$dP =   \frac{2\times 217}{19} (-2.25) + \frac{-217^2}{19^2}(-0.04) $$

ganeshie8 · Answer

thats the usual total differential formula right ?

ganeshie8 · Answer

for $\Delta P$, just find the change in function

ganeshie8 · Answer

$$\Delta P = \dfrac{(217-2.25)^2}{19-0.04} - \dfrac{217^2}{19}$$

ganeshie8 · Answer

Okay for $\Delta P$, you just need to find the change between final and initial values of   P = V^2/R

ganeshie8 · Answer

final (V, R)  :  (217-2.25,  19-0.04)

initial (V, R) : (217,  19)

ganeshie8 · Answer

evaluate the power function at these values and take the difference

ganeshie8 · Answer

np :) sorry to confuse you in the start as i was thinking the question wants only the approximation..