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At what points does 25x^{2} + 36y^{2} = 900 have minimum radius of curvature? Enter the points in...

Question:

At what points does {eq}\displaystyle 25x^{2} + 36y^{2} = 900 {/eq} have minimum radius of curvature?

Enter the points in ascending order of their x -coordinates.

Radius of curvature:

The radius of curvature is defined as the reciprocal of curvature. In other words, the radius of curvature is the radius which is formed with the curve part. The radius of curvature is simply denoted by letter (R). The radius of curvature plays an important role in the field of Physics, Engineering, etc.

Answer and Explanation:


We have given that,

{eq}25x^2 + 36y^2 = 900 {/eq}

Step-1

Now, we will write to above equation in standard form.

{eq}\begin{align*} \dfrac{{25x^2 }}{{900}} + \dfrac{{36y^2 }}{{900}} &= 1 \\ \dfrac{{x^2 }}{{36}} + \dfrac{{y^2 }}{{25}} &= 1 \\ \left( {\dfrac{x}{6}} \right)^2 + \left( {\dfrac{y}{5}} \right)^2 &= 1 \\ \end{align*} {/eq}


We know that,

{eq}\cos ^2 \theta + \sin ^2 \theta = 1 {/eq}

By comparing

{eq}\dfrac{x}{6} = \cos \theta {/eq} and {eq}\dfrac{y}{5} = \sin \theta {/eq}


{eq}x = 6\cos \theta {/eq} and, {eq}y = 5\sin \theta {/eq}


Then, parametric equation is,

{eq}\begin{align*} \overrightarrow r &= \left\langle {x,y} \right\rangle \\ r\left( \theta \right) &= \left\langle {6\cos \theta ,5\sin \theta } \right\rangle \\ \end{align*} {/eq}


Step-2

Now, we will find the first and second derivative of the above equation with respect to {eq}\theta {/eq}.

{eq}\begin{align*} r'\left( \theta \right) &= \left\langle { - 6\sin \theta ,5\cos \theta } \right\rangle \\ r''\left( \theta \right) &= \left\langle { - 6\cos \theta , - 5\sin \theta } \right\rangle \\ \end{align*} {/eq}


Now,

{eq}\begin{align*} \left| {r'\left( \theta \right)} \right| &= \sqrt {\left( { - 6\sin \theta } \right)^2 + \left( {5\cos \theta } \right)} \\ &= \sqrt {36\sin ^2 \theta + 25\cos ^2 \theta } \\ \end{align*} {/eq}


{eq}r'\left( \theta \right) \times r''\left( \theta \right) {/eq} {eq}= \left| {\begin{align*} {\hat i} && {\hat j} \\ { - 6\sin \theta } && {5\cos \theta } \\ { - 6\cos \theta } && { - 5\sin \theta } \\ \end{align*}} \right| {/eq}

{eq}\begin{align*} &= \hat i\left[ {30\sin ^2 \theta } \right] + j\left[ {30\cos ^2 \theta } \right] \\ &= 30\left\langle {\sin ^2 \theta + \cos ^2 \theta } \right\rangle \\ &= 30 \\ \end{align*} {/eq}


Step-3

Curvature {eq}\left( k \right) = \dfrac{{\left| {r'\left( \theta \right) \times r''\left( \theta \right)} \right|}}{{\left| {r'\left( \theta \right)} \right|^3 }} {/eq}

Since, radius of curvature is minimum when curvature is maximum.


So,

{eq}\begin{align*} k &= \dfrac{{30}}{{\left( {36\sin ^2 \theta + 25\cos ^2 \theta } \right)^{\dfrac{3}{2}} }} \\ k &= 30\left[ {36\sin ^2 \theta + 25\cos ^2 \theta } \right]^{ - \dfrac{3}{2}} \\ k'\left( \theta \right) &= 30\left( { - \dfrac{3}{2}} \right)\left[ {36\sin ^2 \theta + 25\cos ^2 \theta } \right]^{ - \dfrac{5}{2}} \left[ {36\left( {2\sin \theta \cos \theta } \right) + 25\left( { - 2\cos \theta \sin \theta } \right)} \right] \\ k'\left( \theta \right) &= - 45\left[ {36\sin ^2 \theta + 25\cos ^2 \theta } \right]^{ - \dfrac{5}{2}} \left[ {72\sin \theta \cos \theta - 50\cos \theta \sin \theta } \right] \\ \end{align*} {/eq}


Put {eq}k'\left( \theta \right) = 0 {/eq}


So,

{eq}\begin{align*} \left[ {72\sin \theta \cos \theta - 50\sin \theta \cos \theta } \right] &= 0 \\ 22\sin \theta \cos \theta &= 0 \\ \theta &= 0,\dfrac{\pi }{2},\pi ,\dfrac{{3\pi }}{2} \\ \end{align*} {/eq}

Step-4

1 {eq}\to {/eq} For {eq}\theta < 0,\;k' > 0 {/eq}

And {eq}\theta > 0,\;k' < 0 {/eq}

So, curvature maximum at {eq}\theta = 0 {/eq}


2 {eq}\to {/eq} for {eq}\theta < \dfrac{\pi }{2},k' < 0 {/eq}

And, {eq}\theta > \dfrac{\pi }{2},k' > 0 {/eq}

So, curvature minimum at {eq}\theta = \dfrac{\pi }{2} {/eq}


3 {eq}\to {/eq} for {eq}\theta < \pi ,\;k' > 0 {/eq}

And, {eq}\theta > \pi ,\;k' < 0 {/eq}

So, curvature maximum at {eq}\theta = \pi {/eq}


4 {eq}\to {/eq} for {eq}\theta < \dfrac{{3\pi }}{2},k' < 0 {/eq}

And, {eq}\theta > \dfrac{{3\pi }}{2},\;k' > 0 {/eq}

So, curvature minimum at {eq}\theta = \dfrac{{3\pi }}{2} {/eq}

Therefore, radius of curvature minimum at {eq}\theta = 0\;{\rm{and}}\;\theta {\rm{ = }}\pi {/eq}


Step-5

Then,

{eq}\begin{align*} x &= 6\cos \theta \\ &= 6\cos 0^ \circ \\ &= 6 \\ \end{align*} {/eq}


And,

{eq}\begin{align*} x &= 6\cos \pi \\ x &= 6\left( { - 1} \right) \\ x &= - 6 \\ \end{align*} {/eq}

Hence, {eq}x = 6, {/eq} {eq}x = - 6 {/eq} are the coordinate where our radius of curvature is minimum.


Learn more about this topic:

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Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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