Question

Consider the path r(t)=(10t,5t2,5lnt) defined for t>0. Find the length of the curve between the points (10,5,0) and (30,45,5ln(3)).

Answer 1

To find the length of the curve between two points, we can use the arc length formula for a parametric curve and integrate.

Step-by-step explanation:

To find the length of the curve between the points (10,5,0) and (30,45,5ln(3)), we can use the arc length formula for a parametric curve:

L = ∫ab √((dx/dt)2 + (dy/dt)2 + (dz/dt)2) dt

For the given curve r(t) = (10t,5t^2,5ln(t)), we can find the derivatives of x, y, and z with respect to t:

dx/dt = 10, dy/dt = 10t, dz/dt = 5/t

Substituting these derivatives into the arc length formula, we get:

L = ∫ab √(10^2 + (10t)^2 + (5/t)^2) dt

Integrating this expression from t = 1 to t = 3, we can find the length of the curve.

Answer 2

The length of the curve
`C` is

`\displaystyle\int_C\mathrm dS=\int_1^3\left|(\mathrm d\vec r(t))/(\mathrm dt)\right|\,\mathrm dt`

`=\displaystyle\int_1^3\sqrt{\left((\mathrm d(10t))/(\mathrm dt)\right)^2+\left((\mathrm d(5t^2))/(\mathrm dt)\right)^2+\left((\mathrm d(5\ln t))/(\mathrm dt)\right)^2}\,\mathrm dt`

`=\displaystyle\int_1^3\sqrt{100+100t^2+(25)/(t^2)}\,\mathrm dt`

`=\displaystyle\int_1^3\sqrt{(25)/(t^2)}√(4t^2+4t^4+1)\,\mathrm dt`

We have
`√(t^2)=|t|`, but integrating over [1, 3] makes
`t>0`, so
`|t|=t`. Under the square root, we can factorize

`4t^4+4t^2+1=(2t^2+1)^2`

and for the same reason as before,
`√((2t^2+1)^2)=2t^2+1`

Then the integral is

`=\displaystyle5\int_1^3\frac{2t^2+1}t\,\mathrm dt`

`=\displaystyle5\int_1^3\left(2t+\frac1t\right)\,\mathrm dt`

`=5(t^2+\ln t)\bigg|_1^3=\boxed{40+5\ln3}`