Question #100159

3. Let L be the length of the curve x=e^y from (1,0) to (e,1), integrating with respect to x an integral expression for L is_____.


4. Let be the length of the curve y=ln(x) from (1,0) to (e,1), integrating with respect to y, an integral expression for L is_____.

Expert's answer

3.

Changing the parameter of the curve to xx we get the expression

y=lnx,  x[1,e].y = \ln x,\; x \in [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.


Answer: L=1e1+x2xdx.L = \int_1^e \frac{\sqrt{1+x^2}}{x} dx.


4.


Similarly, let us change the curve parameter to yy

x=ey,  y[0,1].x = e^y,\;y \in [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.


Answer: L=011+e2ydy.L = \int_0^1 \sqrt{1 + e^{2y}}dy.


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