How can I find the tangent line to y(x) at x=0 using the integral equation?

[Chen]

y(x) is defined by this equation:

$$\int_0^y{\sqrt[3]{1+t^2} dt} + \int_x^0{(2t+1)^2 dt} = 0$$

How do I find the tangent line to y(x) at x=0?

Here's what I tried to do:

$$y(x) = \int_0^y{\sqrt[3]{1+t^2} dt} = \int_0^x{(2t+1)^2 dt}$$

Taking the derivative of both sides of the equation:

$$y'(x) = ... = (2x+1)^2 = 4x^2 + 4x + 1$$
$$y'(x=0) = 1$$

First, is this method correct?

Second, how do I find y(x=0)? I'm guessing that it's 0, but I'm not sure.

Thanks,
Chen

 

[scan]

seems simple enough...

just substitue the first and second integrals using the y and x as if they're values and equate them to each other like you have done. find y'. may have to use implicit differentiation. then equate to 0

 

[HallsofIvy]

You say that y is defined by
$$\int_0^y{\sqrt[3]{1+t^2} dt} + \int_x^0{(2t+1)^2 dt} = 0$$

It does NOT follow that
$$y(x) = \int_0^y{\sqrt[3]{1+t^2} dt} = \int_0^x{(2t+1)^2 dt}$$
There is no reason for the first "y= ". Saying that y is "defined by" the formula does not mean you can just set y equal to both parts- it means that for any given value of x, y must have the value that makes the two integrals equal. In particular, when x= 0, the right hand side is 0 so y must be such that $\int_0^y{\sqrt[3]{1+t^2}dt}= 0$ and that means that y(0)= 0, not 1.

You can use the "fundamental theorem of calculus" to find the derivative of y:

The derivative of the left side is $\sqrt[3]{1+ y^2}y'$ while the the derivative of the right side is -(2x+1)². In particular, at x= 0, y= 0,
$\sqrt[3]{1+0^2}y'(0)= -(2(0)+1)^2$ or y'(0)= -1. The tangent line goes through (0,0) and has slope -1: y= -x.

 

[Chen]

Thanks Ivy.