Question

Find the radius of curvature of the curve x=t2, y=t at t=1

Answer 1

The radius of curvature of the curve $x = t^2$ and y = t at t = 1 is$(3^3/^2 / 2.$

The radius of curvature of a curve is a measure of how sharply the curve is bent at a certain point. To find the radius of curvature of the curve x = t^2 and y = t at t = 1, we can use the following formula:

$1/R = (x''y' - y''x') / (x'^2 + y'^2)^(3/2)$

where R is the radius of curvature, x' and y' are the first derivatives of x and y with respect to t, and x'' and y'' are the second derivatives of x and y with respect to t.

First, we'll find the first derivatives of x and y with respect to t:

x' = 2t

y' = 1

Then, we'll find the second derivatives of x and y with respect to t:

x'' = 2

y'' = 0

Now we can substitute these values into the formula for the radius of curvature:

$1/R = (2 * 1 - 0 * 2t) / (2t^2 + 1)^(3/2)$

And finally, we'll substitute t = 1 into the equation:

$1/R = (2 * 1 - 0 * 2) / (2 + 1)^(3/2)$

$1/R = 2 / 3^(3/2)$

$R = 1 / (2 / 3^(3/2))$

$R = (3^(3/2)) / 2$

So the radius of curvature of the curve x = $t^2$ and y = t at t = 1 is$(3^3/^2)) / 2.$

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