## anonymous 5 years ago use the differentials to approximate [9.02]to power 1.5,plus [1divide by [9.02]raised to power 0. 5]-26

1. anonymous

2. anonymous

Assuming you mean,$(9.02)^{3/2}\left( \frac{1}{(9.02)^{1/2}} \right)^{-26}$you can consider the following. The differential of y (i.e., dy) is approximately equal to the exact change in y (i.e., Δy), provided that the change in x (Δx) is relatively small. Since for small changes in x,$\frac{\Delta y}{\Delta x}\approx \frac{dy}{dx} \rightarrow \Delta y \approx \frac{dy}{dx}\Delta x$you can approximate a small change in a function y as,$y+\Delta y \approx y + \frac{dy}{dx}\Delta x$The function you need to consider is $y=x^{1/2}$since you're looking at combinations involving $9.02^{1/2}$

3. anonymous

For $x=9 \rightarrow (9)^{1/2}=3$which is easy. But you'll notice that x=9.02 is the old x-value plus a *small* change, so we can approximate the actual value of the function by the linear approximation we derived above:$(x+\Delta x)^{1/2} \approx x^{1/2}+ \frac{1}{2}x^{-1/2}\Delta x$Our x-value is 9 and the small change is 0.02. So,$(9+0.02)^{1/2} \approx (9)^{1/2}+\frac{1}{2} (9)^{-1/2}(0.02)=3+\frac{1}{300}=\frac{901}{300}$

4. anonymous

You can now use this result in your equation:$(9.02)^{3/2}\left( \frac{1}{(9.02)^{1/2}} \right)^{-26} \approx \left( \frac{901}{300} \right)^3\left( \frac{1}{\left( \frac{900}{300} \right)} \right)^{-26}$$=\left( \frac{901}{300} \right)^3\left( \frac{901}{300} \right)^{26}=\left( \frac{901}{300} \right)^{29}$

5. anonymous

If you have any difficulty with what I've written, let me know.

6. anonymous

Incidentally, to nine significant figures, your original expression gives,$7.08752811 \times 10^{13}$while the approximate solution found yields,$7.087654705 \times 10^{13}$You can see that they're pretty close.