Help please and thank you

S

sstudent

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Joined
Nov 26, 2005
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Hi there all, I am new to the forum. I am and have been struggling with calculus and general so I figured what better way then to hop on a math forum and seek help. Wondering if someone can show me the answer to this question with the necessary steps required to answer the question. Thanks all!

Find all points (x,y) on the graph of x^2/3 + y^2/3=8 where the lines tangent to the graph at (x,y) have slope -1.
 
This is just a linear equation, a straight line. Either the slope (derivative) is always dy/dx = m = -1, or it never is.

Solve for "y=", and read off the slope.

Eliz.
 
there was an error in my question, I have fixed it and believe it to be a circle now.
 
I apologize about the multi post... I did not realize the 2 forums were connected. i am just stressed as my tutor went home for the holidays, so I am somewhat scrambling. Thanks to all of you that responded, and would it be okay if I requested assistance on one more problem?
 
It's not a matter of any "affiliation", so much as the same folks volunteering in many places. It can be a bit aggravating to reply to a question on one forum, just to find it later re-posted elsewhere, as though it hadn't already been answered. It can lead tutors to conclude that the student isn't looking for help, but is wanting the complete worked solution to copy into his homework. Just FYI....

Meanwhile, regarding your revised equation, do you mean "x<sup>2</sup>/3 + y<sup>2</sup>/3 = 8" (in other words, x<sup>2</sup> + y<sup>2</sup> = 24), or "x<sup>2/3</sup> + y<sup>2/3</sup> = 8"?

To post a new question, please start a new thread. Thank you.

Eliz.
 
sstudent said:
I was intending the post to read the ladder of your post.
Click to expand...
:shock:

My post has a ladder...? Your post can read...?

Eliz.
 
my question reads as follows... I am sorry for the mispost originally.

Find all points on (x,y) on the graph of x^(2/3)+y^(2/3)=8 where the lines tangent to the graph at (x,y) have slope -1.
 
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