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

In summary, the conversation discusses finding the tangent line to y(x) at x=0 using the equation y(x) = \int_0^y{\sqrt[3]{1+t^2} dt} + \int_x^0{(2t+1)^2 dt} = 0 and finding the value of y(x=0). The method of taking the derivative and using implicit differentiation is suggested. It is also clarified that y(x) cannot be set equal to both parts of the equation and that y(0) is 0, not 1. Finally, the derivative of y is found to be -1 at x=0, giving the equation of the tangent line as y= -x.
  • #1
Chen
977
1
y(x) is defined by this equation:

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

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

Here's what I tried to do:

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

Taking the derivative of both sides of the equation:

[tex]y'(x) = ... = (2x+1)^2 = 4x^2 + 4x + 1[/tex]
[tex]y'(x=0) = 1[/tex]

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
 
Physics news on Phys.org
  • #2
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
 
  • #3
You say that y is defined by
[tex]\int_0^y{\sqrt[3]{1+t^2} dt} + \int_x^0{(2t+1)^2 dt} = 0[/tex]

It does NOT follow that
[tex]y(x) = \int_0^y{\sqrt[3]{1+t^2} dt} = \int_0^x{(2t+1)^2 dt}[/tex]
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 [itex]\int_0^y{\sqrt[3]{1+t^2}dt}= 0[/itex] 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 [itex]\sqrt[3]{1+ y^2}y'[/itex] while the the derivative of the right side is -(2x+1)2. In particular, at x= 0, y= 0,
[itex]\sqrt[3]{1+0^2}y'(0)= -(2(0)+1)^2[/itex] or y'(0)= -1. The tangent line goes through (0,0) and has slope -1: y= -x.
 
  • #4
Thanks Ivy. :smile:
 

1. What is an integral in calculus?

An integral in calculus is a mathematical concept that represents the area under a curve on a graph. It is a fundamental tool in calculus for calculating the total change in a function over a given interval.

2. How is an integral different from a derivative?

While a derivative represents the instantaneous rate of change of a function at a specific point, an integral represents the accumulation of change over a specific interval. In other words, a derivative tells us how a function is changing at a given point, while an integral tells us how much the function has changed over a given interval.

3. What is the process for solving an integral?

The process for solving an integral involves finding an antiderivative of the function being integrated and evaluating it at the upper and lower limits of the given interval. This is typically done using integration techniques such as u-substitution, integration by parts, or partial fractions.

4. What are some real-life applications of integrals?

Integrals have numerous real-life applications in fields such as physics, engineering, economics, and statistics. For example, in physics, integrals are used to calculate the work done by a force, while in economics, they are used to calculate total revenue and profit. In statistics, integrals are used to calculate probabilities and to find areas under probability distributions.

5. How can I improve my skills in solving integrals?

The best way to improve your skills in solving integrals is to practice regularly. Start by understanding the basic concepts and techniques, and then move on to more advanced problems. You can also seek help from a tutor or use online resources such as practice problems and video tutorials to supplement your learning.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
670
  • Introductory Physics Homework Help
2
Replies
40
Views
779
  • Introductory Physics Homework Help
Replies
7
Views
865
  • Introductory Physics Homework Help
Replies
11
Views
914
  • Introductory Physics Homework Help
2
Replies
64
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
863
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
282
  • Calculus and Beyond Homework Help
Replies
2
Views
260
  • Introductory Physics Homework Help
Replies
1
Views
144
Back
Top