MHB Linear Approximation With Trigonometry

[ardentmed]

Hey guys,

I have just a few more questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

I'm only asking about 1ab, ignore 2abc please:

So for the first one, I used cos(30) as the estimated value to approximate L(28). Then I converted it to radians to get $\pi$/6

Knowing that f'(x) = -sin(x), I calculated L(x):

L(x) = √ (3)/2 + (1/2)(x- $\pi$/6) which ultimately gave me 0.85 for L(28). I'm highly doubtful that this is the right answer though as there may be an exact value response which I may have missed.

As for 1b, I took x=0 this time since e^(-0.00015) is close to one, similarly to e^0 . As such, I took L(x) with that estimation and got:

L(-0.0015) = 1,997/2,000Thanks in advance.

 

[MarkFL]

1a) Yes, converting to radians is what you want to do here. For small $\Delta x$, we may state:

$$\frac{\Delta f}{\Delta x}\approx\d{f}{x}$$

$$\Delta f\approx\d{f}{x}\Delta x$$

$$f\left(x+\Delta x\right)-f(x)\approx\d{f}{x}\Delta x$$

$$f\left(x+\Delta x\right)\approx\d{f}{x}\Delta x+f(x)$$

Now, if we define:

$$f(x)\equiv\cos(x)\implies f^{\prime}(x)=-\sin(x)$$

we obtain:

$$\cos\left(x+\Delta x\right)\approx-\sin(x)\Delta x+\cos(x)$$

Now, letting:

$$x=\frac{\pi}{6},\,\Delta x=-\frac{\pi}{90}$$

we obtain:

$$\cos\left(\frac{\pi}{6}-\frac{\pi}{90}\right)\approx-\sin\left(\frac{\pi}{6}\right)\cdot\frac{\pi}{90}+\cos\left(\frac{\pi}{6}\right)$$

Hence:

$$\cos\left(28^{\circ}\right)\approx\frac{\sqrt{3}}{2}-\frac{\pi}{180}$$

Your value is correct to two decimal places.

1b) Using the same method, we may begin with:

$$f\left(x+\Delta x\right)\approx\d{f}{x}\Delta x+f(x)$$

If we define:

$$f(x)\equiv e^x\implies f^{\prime}=f(x)$$

we obtain:

$$e^{x+\Delta x}\approx e^x\Delta x+e^x$$

Now, letting:

$$x=0,\,\Delta x=-0.0015$$

we obtain:

$$e^{-0.0015}\approx1-0.0015=\frac{1997}{2000}$$

So, our answers agree. :D