Question

37-38= Use the formula in Exercise 36 to find the curvature.s

37. \(x = {e^t}cost,\;\;\;y = {e^t}sint\)

Short answer

The curvature for the given curve is, \(\kappa = \frac{{{e^{2t}}}}{{\sqrt 2 {e^{3t}}}}\)

Step-by-step solution

Step 1: To find the curvature of the curve

According to formula of curvature, we know that,

\(\kappa = \frac{{\left| {{x^'}{y^{''}} - {y^'}{x^{''}}} \right|}}{{{{\left( {{{\left( {{x^'}} \right)}^2} + {{\left( {{y^'}} \right)}^2}} \right)}^{\frac{3}{2}}}}}\)

The parametric equation for the given curve is,

\(\left\langle {x = {e^t}cost,y = {e^t}sint} \right\rangle \)

Differentiating the above equation with respect to time \(t,\) we get,

\(\left\langle {{x^'} = {e^t}cost + {e^t}( - sint),{y^'} = {e^t}sint + {e^t}cost} \right\rangle \)

that is,

\(\left\langle {{x^'} = {e^t}(cost - sint),{y^'} = {e^t}(sint + cost)} \right\rangle \)

Similarly, we could find the second derivative, that is,

\(\left\langle {{x^{''}} = {e^t}(cost - sint) + {e^t}( - sint - cost),{y^{''}} = {e^t}(sint + cost) + {e^t}(cost - sint)} \right\rangle \)

that is,

\(\left\langle {{x^{''}} = - 2{e^t}sint,{y^{''}} = 2{e^t}cost} \right\rangle \)

Step 2: Final proof

Applying the values of \({x^'},{y^'},{x^{''}}\) and \({y^{''}}\) into the formula for curvature, we get,

\(\kappa = \frac{{\left| {{x^'}{y^{''}} - {y^'}{x^{''}}} \right|}}{{{{\left( {{{\left( {{x^'}} \right)}^2} + {{\left( {{y^'}} \right)}^2}} \right)}^{\frac{3}{2}}}}}\)

\(\kappa = \frac{{\left| {{e^t}(cost - sint)\left( {2{e^t}cost} \right) - {e^t}(sint + cost)\left( { - 2{e^t}sint} \right)} \right|}}{{{{\left( {{{\left( {{e^t}(cost - sint)} \right)}^2} + {{\left( {{e^t}(sint + cost)} \right)}^2}} \right)}^{\frac{3}{2}}}}}\)

\(\kappa = \frac{{\left| {2{e^{2t}}\left( {co{s^2}t - costsint} \right) + 2{e^{2t}}\left( {si{n^2}t + costsint} \right)} \right|}}{{{{\left( {{e^{2t}}(1 - 2costsint) + {e^{2t}}(1 + 2costsint)} \right)}^{\frac{3}{2}}}}}\)

\(\kappa = \frac{{\left| {2{e^{2t}}\left( {co{s^2}t + si{n^2}t} \right)} \right|}}{{{2^{\frac{3}{2}}}{e^{3t}}}}\)

\(\kappa = \frac{{\left| {{e^{2t}}} \right|}}{{\sqrt 2 {e^{3t}}}}\) \((since\left. {{{\cos }^2}\theta + {{\sin }^2}\theta = 1} \right)\)