Answered: 8x^2y^2 = (x^2 + y^2 )^3 List one…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Concept explainers

Question

8x^2y^2 = (x^2 + y^2 )^3

List one point on the curve at which the tangent line is horizontal. Show all work

Expert Solution

Step 1

The given curve is $8 x^{2} y^{2} = {(x^{2} + y^{2})}^{3}$ …(1)

Differentiating the given function with respect to x we get

$\begin{array}{rcl} 16 x y^{2} + 16 x y^{2} y' & = & 3 {(x^{2} + y^{2})}^{2} . (2 x + 2 y y') \\ 16 x y^{2} y' - 6 y {(x^{2} + y^{2})}^{2} y' & = & 6 x {(x^{2} + y^{2})}^{2} - 16 x y^{2} \\ y' (16 x y^{2} - 6 y {(x^{2} + y^{2})}^{2}) & = & 6 x {(x^{2} + y^{2})}^{2} - 16 x y^{2} \\ y' & = & \frac{6 x {(x^{2} + y^{2})}^{2} - 16 x y^{2}}{16 x y^{2} - 6 y {(x^{2} + y^{2})}^{2}} \end{array}$

The tangent line will be horizontal only when slope is 0.

i.e. $y' = 0$ which gives

Step by step

Solved in 2 steps

Knowledge Booster

$Single Variable$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions