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Using differential find approximate value of $\sqrt[3]{1.02}$

I did this. We know $f(x_0+\Delta x)-f(x_0)=\frac{df}{dx}(x_0)\cdot\Delta x$

$$f(x)=\sqrt[3]{1+x}$$

$$f(0.02)-f(0)=\frac{df}{dx}(0)\cdot0.02$$

And after calculation I solved.

But the problem I don't know how to solve is approximate using differential $\sin29^\circ$.

Michael Hardy
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    do the same but for the sin function. sin(a+x), when a is a value of sin you know. remember $sin(x+2\pi)=sin(x)$ – Orenio May 02 '21 at 17:09
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    The title of the question does not match the body. Can you fix this? – Pedro May 02 '21 at 17:20

1 Answers1

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The value of $\sin (30^\circ) = \sin(\frac\pi6) = 0.5$ is known.

Now you have to convert the angles to radians.

Let $x_0 = \frac\pi6$, $\Delta x = - 1^\circ (\text{in radian}) =- \frac{\pi}{180} $.

And $\frac{df}{dx}(x_0) = \cos(x)_{x_0= \pi/6} $

$\sin(29^\circ)=f(x+x_0) = f(x_0)+\frac{df}{dx}(x_0)\Delta x = \sin\frac\pi6 - (\cos\frac\pi6)\cdot\frac\pi{180} \approx 0.4848$

P.S. This question has answers for why we should convert degrees to radians in calculus

19aksh
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  • I am getting $\frac{1-\sqrt{3}}{2}$ which is not the answer. –  May 02 '21 at 17:24
  • That's what I've mentioned, you should use $\boxed{\Delta x = - \frac{\pi}{180}}$. I think you've used $\Delta x = -1$ which is in degrees.. – 19aksh May 02 '21 at 17:27
  • $f(x)=sin(\frac{\pi}{6}+x)$ . $f(\frac{-pi}{180})-f(0)=\frac{df}{dx}(0)*\frac{-pi}{180}$ from here not getting the answer. –  May 02 '21 at 17:43
  • @rha I've added it in the answer – 19aksh May 02 '21 at 17:48
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    Thanks you very much for your kindness!. –  May 02 '21 at 17:52