Write an equation for the tangent to the curve y=sinmx at the origin.
I got y=mx, is this correct? Please help by showing work! Thanks
correct.
If you got that, I think your work must have been fine. It's due to the chain rule.
To find the equation for the tangent to the curve y = sin(mx) at the origin, we need to find the derivative of the given function and then evaluate it at x = 0.
Step 1: Find the derivative of y = sin(mx) with respect to x using the chain rule.
Let u = mx
dy/dx = d(sin(u))/du * du/dx
Since sin(u) has a derivative of cos(u) with respect to u, and du/dx = m, we can substitute these values to get:
dy/dx = cos(mx) * m
Step 2: Evaluate the derivative at x = 0 to get the slope of the tangent line.
dy/dx = cos(m * 0) * m
dy/dx = cos(0) * m
dy/dx = m
Step 3: Find the equation of the tangent line using the slope-intercept form.
Since the tangent line passes through the origin (0, 0), the y-intercept (b) is 0.
Therefore, the equation of the tangent line is:
y = m * x
So, your answer of y = mx is correct!
To find the equation for the tangent to the curve y = sin(mx) at the origin, we need to first find the derivative of the function with respect to x.
Let's start by finding the derivative:
dy/dx = d/dx(sin(mx))
To differentiate sin(mx), we can use the chain rule. The derivative of sin(mx) with respect to x is equal to the derivative of sin(u) multiplied by the derivative of mx, where u = mx.
dy/dx = cos(mx) * m
Now, we can evaluate the derivative at the origin (x = 0) to find the slope of the tangent line:
dy/dx = cos(0) * m
dy/dx = 1 * m
dy/dx = m
Therefore, the slope of the tangent line is equal to m.
To find the equation of the tangent line passing through the origin (0,0), we can use the point-slope form of the equation:
y - y1 = m(x - x1)
Since the point on the line is (0,0), we have:
y - 0 = m(x - 0)
y = mx
Thus, the equation of the tangent line to the curve y = sin(mx) at the origin is y = mx. So, your answer is correct!