utkarshakash said: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.
The slope of a curve at x=0 represents the instantaneous rate of change of the curve at that specific point. It tells us how much the y-value is changing for every unit change in the x-value at that point.
To find the slope of a curve at x=0, we can use the derivative of the function y(x). We can take the derivative with respect to x and then substitute x=0 into the resulting equation to find the slope at that point.
Yes, the slope of a curve at x=0 can be negative. This means that the curve is decreasing at that point, and the y-value is decreasing for every unit increase in the x-value.
A slope of 0 at x=0 indicates that the curve is horizontal at that point. This means that the y-value is not changing as the x-value increases or decreases.
Finding the slope of a curve at x=0 is important because it allows us to understand the behavior of the curve at that specific point. It can also help us determine the direction of the curve and make predictions about its future behavior.