tangents are drawn from the origin to the curve y=sinx, then their point of contact lies on the curve _________
dy/dx = cosx
at x = 0 , cos 0 = 1
so the slope of the tangent is 1 and the y-intercept is 0 (the origin)
tangent equation :
y = x
To determine the point of contact between the tangents and the curve y = sin(x), we need to find the equation of the tangents.
The derivative of the curve y = sin(x) is given by dy/dx = cos(x). This represents the slope of the curve at any given point.
Since the tangents are drawn from the origin (0, 0), the slope of the line passing through the origin and any point (x, sin(x)) on the curve is simply:
slope = (sin(x) - 0)/(x - 0) = sin(x)/x
We need to find the values of x for which the slope of the tangents is equal to sin(x)/x.
Using trigonometric identities, we know that sin(0) = 0, so sin(x)/x = 0 when x = 0.
Thus, the point of contact between the tangents and the curve y = sin(x) lies on the curve when x = 0.
To find the point of contact where the tangents are drawn from the origin to the curve y = sin(x), we need to determine the equation of the curve at that point.
Let's break down the process step by step:
1. Find the derivative of the curve y = sin(x) using the chain rule:
dy/dx = cos(x)
2. The derivative dy/dx represents the slope of the curve at any point on the curve.
3. Since the tangents are drawn from the origin (0, 0), the slope of the tangents will be equal to the slope of the curve at the point of contact.
4. The slope of the tangent is given by the ratio of the y-coordinate to the x-coordinate of the point of contact. Since the tangents are drawn from the origin (0, 0), the slope is simply y/x.
5. Setting the slope of the curve equal to y/x, we have:
cos(x) = y/x
6. Rearranging the equation, we get:
y = x * cos(x)
Therefore, the point of contact where the tangents are drawn from the origin to the curve y = sin(x) lies on the curve y = x * cos(x).