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Find the slope of the tangent line to the curve (a lemniscate) 2(x^{2}+y^{2})^{2}=25(x^{2}-y^{2})

Question:

Find the slope of the tangent line to the curve (a lemniscate)

{eq}2(x^{2}+y^{2})^{2}=25(x^{2}-y^{2}) {/eq} at the point {eq}(3, 1) {/eq}.

Answer and Explanation:

Consider the function

{eq}\displaystyle2(x^{2}+y^{2})^{2}=25(x^{2}-y^{2}),\quad ( 3,1) {/eq}

Differentiate with respect to {eq}x {/eq} on both sides

{eq}\displaystyle4(x^{2}+y^{2})\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}+y^{2})=25\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}-y^{2})\\ \displaystyle4(x^{2}+y^{2})\left ( 2x+2y\frac{\mathrm{d} y}{\mathrm{d} x} \right )=25\left ( 2x-2y\frac{\mathrm{d} y}{\mathrm{d} x} \right )\\ \displaystyle4(x^{2}+y^{2})\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )=25\left ( x-y\frac{\mathrm{d} y}{\mathrm{d} x} \right )\\ \displaystyle4x(x^{2}+y^{2})+4y(x^{2}+y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}=25x-25y\frac{\mathrm{d} y}{\mathrm{d} x}\\ \displaystyle(4y(x^{2}+y^{2})+25y)\frac{\mathrm{d} y}{\mathrm{d} x}=25x-4x(x^{2}+y^{2})\\ \displaystyle\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{25x-4x(x^{2}+y^{2})}{4y(x^{2}+y^{2})+25y} {/eq}

Hence, the slope at point (3,1) is as follows

{eq}\displaystyle m=\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right) \Biggr|_{(3,1)}=\frac{25(3)-4(3)(3^{2}+1^{2})}{4(1)(3^{2}+1^{2})+25(1)}=-\frac{9}{13}\\ {/eq}


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