Question

find the point on the curve y=x^2 where the slope of the tangent is equal to the x-coordinate of the point.

Answer 1

What have you tried? This is a fun one! =)

Answer 2

i found the slope dy/dx = 2x ... then what ??

Answer 3

So what's the slope equal to? It says it's equal at a point. Maybe there's some way you can plug in numbers or equate things?

Answer 4

Maybe drawing it out and labelling where approximately you think that point might be.

Answer 5

we have to solve it using derivatives

Answer 6

Read through the question piece by piece and see if you can understand what it's asking.

Answer 7

if you know how to solve the problem then please show it .. else don't waste time !!

Answer 8

I'm not. I'm helping you more than you realize. Think about the question and try to understand. If I show you the answer it would be giving it away. You need to give it an honest attempt. I will help you if you try.

Answer 9

i have read the sum several times, tried doing it on my own ... when i couldn't i posted it here ....

Answer 10

"find the point on the curve y=x^2 where the slope of the tangent is equal to the x-coordinate of the point. "

So if the slope of the tangent is equal to the x-coordinate of the point, what does this mean?

Break it down into pieces.

"slope of the tangent" is the derivative of y, right?
\[y'\]
"is equal to" So we need to equate y' with something
\[y'=\]
"equal to the x-coordinate" So we set it equal to x.\[y'=x\]
So what is y'? it's 2x, so we plug it in to solve for x.
\[2x=x\]
What value of x is this true? There's only one point, and that's the point we're looking for.

Once we find this value, we plug it into y=x^2 to find the coordinates of the point so we can answer the question with (x,y) as our answer.

Answer 11

the point is (0,0) .. .thanks @Kainui !!

Answer 12

Exactly. Does that make sense too? The slope is 0 at the point (0,0) so it even makes sense when you think about it! Relieving!