3.
Changing the parameter of the curve to xxx we get the expression
y=lnx, x∈[1,e].y = \ln x,\; x \in [1,e].y=lnx,x∈[1,e].
Applying the formula of function graph arc length we get
L=∫1e1+(ln′(x))2dx=∫1e1+1x2dx=∫1e1+x2xdx.L = \int_1^e \sqrt{1 + (\ln'(x))^2} dx = \int_1^e \sqrt{1 + \frac{1}{x^2}} dx \\= \int_1^e \frac{\sqrt{1+x^2}}{x} dx.L=∫1e1+(ln′(x))2dx=∫1e1+x21dx=∫1ex1+x2dx.
Answer: L=∫1e1+x2xdx.L = \int_1^e \frac{\sqrt{1+x^2}}{x} dx.L=∫1ex1+x2dx.
4.
Similarly, let us change the curve parameter to yyy
x=ey, y∈[0,1].x = e^y,\;y \in [0,1].x=ey,y∈[0,1].
Then,
L=∫011+(ddyey)2dy=∫011+e2ydy.L = \int_0^1 \sqrt{1 + (\frac{d}{dy}e^y)^2}dy = \int_0^1 \sqrt{1 + e^{2y}}dy.L=∫011+(dydey)2dy=∫011+e2ydy.
Answer: L=∫011+e2ydy.L = \int_0^1 \sqrt{1 + e^{2y}}dy.L=∫011+e2ydy.
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