Slope of the curve

  • #1
utkarshakash
Gold Member
855
13

Homework Statement


Slope of the curve at the point x=0 of the function y=y(x) specified implicitly as [itex]\displaystyle \int_0^y e^{-t^2} dt + \int_0^x \cos t^2 dt = 0[/itex] is

Homework Equations



The Attempt at a Solution



Differentiating both sides wrt x
[itex]e^{-y^2} \frac{dy}{dx} + \cos x^2 = 0[/itex]
Now If I put x=0 above
[itex]\dfrac{dy}{dx} = \dfrac{-1}{e^{-y^2}} [/itex]

But I don't know the value of y when x = 0.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619

Homework Statement


Slope of the curve at the point x=0 of the function y=y(x) specified implicitly as [itex]\displaystyle \int_0^y e^{-t^2} dt + \int_0^x \cos t^2 dt = 0[/itex] is

Homework Equations



The Attempt at a Solution



Differentiating both sides wrt x
[itex]e^{-y^2} \frac{dy}{dx} + \cos x^2 = 0[/itex]
Now If I put x=0 above
[itex]\dfrac{dy}{dx} = \dfrac{-1}{e^{-y^2}} [/itex]

But I don't know the value of y when x = 0.

You should be able to figure it out. Put x=0 into your original integral equation.
 

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