Answer to Question #92549 in Calculus for jinho

Question

1. Evaluate lim 1+ e raised to 2x / xcosx
c approaching 0

2. Apply L Hopital's rule to show that lim ( 1+x ) raised to 1/x = e

3. Find dy/ dx and d squared y / d x raised to 2 given x= sinht, y= casht.

4. Approximate the value of arcsin ( 0.48) by differential.

5. The measured radius of the ballbearing is 0.7 inch. If the measurement is correct to within 0.01 inch, estimate the relative error in the volume of the ball bearing.

Answer 1

1

$$\lim_{x\to 0} (1+e)^{\frac{2x}{xcosx}}=[(1+e)^\frac{0}{0}] =\lim_{x\to 0} (1+e)^{\frac{2}{cosx}}=[(1+e)^\frac{2}{1}]=(1+e)^2$$

2

$$L=\lim_{x\to 0} (1+x)^{\frac{1}{x}}$$

$$\log L=\log\left(\lim_{x\to 0} (1+x)^{\frac{1}{x}}\right)= \lim_{x\to 0} \log(1+x)^{\frac{1}{x}}=\lim_{x\to 0}\frac{\log(1+x)}{x}=$$

$$=[\frac{0}{0}]=\lim_{x\to 0} \frac{\frac{d}{dx}\log(1+x)}{\frac{d}{dx}x}=\lim_{x\to 0} \frac{\frac{1}{1+x}}{1}=1$$ $$L=e$$

3

$$x=sinh(t);\quad y=cosh(t)$$ $$\frac{dy}{dx}=y'=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{sinh(t)}{cosh(t)}=tanh(t)$$ $$\frac{d^2y}{dx^2}=\frac{\frac{dy'}{dt}}{\frac{dx}{dt}}= \frac{\frac{d}{dt}tanh(t)}{\frac{d}{dt}sinh(t)}=\frac{1-tanh^2(t)}{cosh(t)}$$

4.

$$dy=f'(x)(x-x_0)$$ $$f(x)-f(x_0)=df=f'(x)(x-x_0)$$

Let x=0.5 and $x_0=0.48$

$$f(0.5)=\arcsin0.5=\frac{\pi}{6}$$

$$f'(x)=\frac{1}{\sqrt{1-x^2}}$$ $$f'(0.5)=\frac{1}{\sqrt{1-0.25}}=\frac{2}{\sqrt{3}}$$

Hence

$$f(x_0)=f(0.48)=f(0.5)-f'(0.5)*0.02=\frac{\pi}{6}-\frac{0.04}{\sqrt{3}}$$

5.

$$V=\frac{4}{3}\pi r^3$$

$$dV(r)=V'(r)dr$$

r=0.7

dr=0.01

$$V'=4\pi r^2$$ $$\Delta V\approx dV=4\pi (0.7)^2*(\pm0.01)=4\pi*0.49*(\pm0.01)$$

Relative error=$\frac{dV}{V}=\frac{4\pi*0.49*(\pm0.01)}{\frac{4}{3}\pi (0.7)^3}=3*\frac{\pm0.01}{0.7}=\pm0.04$

Or

$$\pm4\%$$