How to solve a²y = x⁴

April 8, 2020 - 2:34pm

#1
Francis June E....

Minima maxima: a²y = x⁴

How to solve a²y = x⁴

Tags:

- 317 reads

Subscribe to MATHalino on

- Tapered Beam
- Vickers hardness: Distance between indentations
- Time rates
- Minima Maxima: y=ax³+bx²+cx+d
- Make the curve y=ax³+bx²+cx+d have a critical point at (0,-2) and also be a tangent to the line 3x+y+3=0 at (-1,0).
- Minima maxima: Arbitrary constants for a cubic
- Minima Maxima: 9a³y=x(4a-x)³
- Minima maxima: a²y = x⁴
- how to find the distance when calculating moment of force
- strength of materials
- Analytic Geometry Problem Set [Locked: Multiple Questions]
- Equation of circle tangent to two lines and passing through a point
- Product of Areas of Three Dissimilar Right Triangles
- Perimeter of Right Triangle by Tangents
- Differential equations
- Laplace
- Families of Curves: family of circles with center on the line y= -x and passing through the origin
- Family of Plane Curves
- Differential equation
- Differential equation

Home • Forums • Blogs • Glossary • Recent

About • Contact us • Disclaimer • Privacy Policy • Hosted by WebFaction • Powered by Drupal

About • Contact us • Disclaimer • Privacy Policy • Hosted by WebFaction • Powered by Drupal

Forum posts (unless otherwise specified) licensed under a Creative Commons Licence.

All trademarks and copyrights on this page are owned by their respective owners. Forum posts are owned by the individual posters.

All trademarks and copyrights on this page are owned by their respective owners. Forum posts are owned by the individual posters.

Your question don't have enough details. What are you trying to solve from this equation? Tagging it with Maxima and Minima is not enough information.

Yung Ans. Niya po is (0,0) at Minimum.

Yung process sana po sir. Thanks po

Yung process po sana kung paano solve.

Yung ans nya po is (0,0) , minimum.

$a^2 y = x^4$

Differentiate

$a^2 y' = 4x^3$

Equate

y'= 0 to determine the critical points (maxima or minima)$a^2 (0) = 4x^3$

$x = 0$

For

x= 0$a^2 y = 0^4$

$y = 0$

Hence,

critical point = (0, 0)

Check the neighboring points to determine whether (0, 0) is minimum or maximum. Set

x= ±1:$a^2 y = (\pm 1)^4$

$y = +\dfrac{1}{a^2}$ ← above (0, 0)

Hence, the point (0, 0) is minimum.

Salamat po sir

How about 9a²y=x(4a+x)³

Yung ans nya po is (a,3a) maximum

Thanks in advance sir

Please create another forum post for your 2nd question as moderators won't allow multiple questions in one thread.