- #1
erik05
- 50
- 0
Just wondering if anyone here could check to see if I did this question right.
An astroid has the equation [tex] x^\frac{2}{3} + y^\frac{2}{3}=1 [/tex]. The equation defines two continuous functions y=y(x) in the interval [tex] -1\leq x \leq 1 [/tex]
a) Solve the equation for y to obtain formulas for the functions.
b) Give formulas for the first derivatives of the functions for x cannot equal 0, ±1 by differentiating the equation of the curve implicitly and then substituting in the formulas for y from part (a).
c) Compute the same derivatives by differentiating the formulas from part (a)
a) [tex] y= (1-x^\frac{2}{3})^\frac{3}{2}[/tex]
[tex] =\pm \sqrt{(1-x^\frac{2}{3})^3} [/tex]
b) [tex] y'= \frac {-\sqrt[3]{y}}{\sqrt[3]{x}} [/tex]
substituting y:
c) [tex] y= (1-x^\frac{2}{3})^\frac{3}{2}) [/tex]
[tex] y'= \frac {3}{2} (1-x^\frac{2}{3})^\frac{1}{2} (\frac{-2}{3}x^\frac{-1}{3}) [/tex]
[tex] = \frac{-1}{\sqrt[3]{x}} (1-x^\frac{2}{3})^\frac{1}{2} [/tex]
Thank you all in advance.
Edit: Sorry if my work is incomplete, it's not letting me post the rest for some reason.
An astroid has the equation [tex] x^\frac{2}{3} + y^\frac{2}{3}=1 [/tex]. The equation defines two continuous functions y=y(x) in the interval [tex] -1\leq x \leq 1 [/tex]
a) Solve the equation for y to obtain formulas for the functions.
b) Give formulas for the first derivatives of the functions for x cannot equal 0, ±1 by differentiating the equation of the curve implicitly and then substituting in the formulas for y from part (a).
c) Compute the same derivatives by differentiating the formulas from part (a)
a) [tex] y= (1-x^\frac{2}{3})^\frac{3}{2}[/tex]
[tex] =\pm \sqrt{(1-x^\frac{2}{3})^3} [/tex]
b) [tex] y'= \frac {-\sqrt[3]{y}}{\sqrt[3]{x}} [/tex]
substituting y:
c) [tex] y= (1-x^\frac{2}{3})^\frac{3}{2}) [/tex]
[tex] y'= \frac {3}{2} (1-x^\frac{2}{3})^\frac{1}{2} (\frac{-2}{3}x^\frac{-1}{3}) [/tex]
[tex] = \frac{-1}{\sqrt[3]{x}} (1-x^\frac{2}{3})^\frac{1}{2} [/tex]
Thank you all in advance.
Edit: Sorry if my work is incomplete, it's not letting me post the rest for some reason.
Last edited: