anonymous
  • anonymous
Find the function y = y(x) such that the line tangent to the graph of y = y(x) at (x, y) intersects the x-axis at x/2 .
Differential Equations
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
i know i have to end up with dy/dx = something but i cant figure out how to get there with this clue... any suggestions?
wolfe8
  • wolfe8
Well, dy/dx is the gradient at a specific point of a curve.
anonymous
  • anonymous
ok, so i need to find the equation of the tangent line

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

dumbcow
  • dumbcow
tangent line \[y-y_1 = m(x-x_1)\] \[m = \frac{dy}{dx}\] line crosses point (x/2 , 0) \[-y_1 = \frac{dy}{dx}(\frac{x_1}{2}-x_1)\] from here you can set up diff equ and separate variables to solve for y(x)
anonymous
  • anonymous
Got you, I see where im going now. The only thing im not sure about is the sign on y. Are u positive is negative?
anonymous
  • anonymous
im meant y being negative...
dumbcow
  • dumbcow
yes because the line is tangent to curve at (x_1, y_1) the "y" value of line is 0 at x intercept 0 - y_1 = -y_1
anonymous
  • anonymous
Got it. So i solved it and got: \[y(x) = x^2 +C\] What do you think?
dumbcow
  • dumbcow
yep well close, i get constant in front y = Cx^2
anonymous
  • anonymous
ahhhhh, yes. Because 2lnx + C becomes the exponent for e and e^C becomes C
dumbcow
  • dumbcow
haha yeah was just going to say that :)
anonymous
  • anonymous
I always mess up whenever e or ln are involved
anonymous
  • anonymous
Thank you for your help, you got me out of a pickle :)
dumbcow
  • dumbcow
no problem

Looking for something else?

Not the answer you are looking for? Search for more explanations.