Find an equation of the tangent line to the curve at the given point.
y = 6e^x cos x, (0, 6)
Why did the tangent line go to the party alone? Because it didn't want any curves to tag along!
To find the equation of the tangent line to the curve y = 6e^x cos(x) at the point (0,6), we first need to find the slope of the tangent line at that point. We can do this by taking the derivative of the function.
The derivative of y with respect to x is given by:
dy/dx = (d/dx)(6e^x cos(x))
Now, let's find the derivative using the product rule:
dy/dx = 6(e^x)(-sin(x)) + 6e^x(-sin(x)) + 6e^x(cos(x))
Simplifying this expression, we get:
dy/dx = -6e^x sin(x) - 6e^x sin(x) + 6e^x cos(x)
At x = 0 (the given point), the derivative becomes:
dy/dx = -6(0) sin(0) - 6(0) sin(0) + 6(1) cos(0) = 6
So, the slope of the tangent line at (0,6) is 6. We can now use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values of (x1, y1) = (0,6) and m = 6, we have:
y - 6 = 6(x - 0)
Simplifying further:
y - 6 = 6x
Finally, rearranging the equation in the slope-intercept form, we get the equation of the tangent line:
y = 6x + 6
So, the equation of the tangent line to the curve y = 6e^x cos(x) at the point (0,6) is y = 6x + 6. See? Tangent lines can be pretty straightforward, unlike those curvy ones!
To find the equation of the tangent line to a curve at a given point, we need to find the slope of the tangent line and then use the point-slope form of a linear equation.
Step 1: Find the derivative of the function y = 6e^x cos x.
To find the derivative, we use the product rule and chain rule:
y' = (6e^x * cos x)' + (6e^x * cos x)'
= (6e^x * cos x)' + (6e^x * (-sin x))
Using the product rule again on the first term:
y' = (6e^x * cos x)' + (6e^x * (-sin x))
= (6e^x)' * cos x + 6e^x * (cos x)' + (6e^x * (-sin x))
= (6e^x) * cos x + 6e^x * (-sin x) + (6e^x * (-sin x))
= 6e^x * cos x - 6e^x * sin x - 6e^x * sin x
Simplifying the expression:
y' = 6e^x * cos x - 12e^x * sin x
Step 2: Substitute the x-coordinate of the given point into the derivative to find the slope.
Substituting x=0 into the derivative:
y' = 6e^0 * cos 0 - 12e^0 * sin 0
= 6 * 1 * 1 - 12 * 0 * 0
= 6
The slope of the tangent line at point (0, 6) is 6.
Step 3: Use the slope and the given point to write the equation of the tangent line using the point-slope form.
Using the point-slope form: y - y1 = m(x - x1)
Substituting the coordinates (x1, y1) = (0, 6) and slope m = 6:
y - 6 = 6(x - 0)
y - 6 = 6x
y = 6x + 6
Therefore, the equation of the tangent line to the curve y = 6e^x cos x at the point (0, 6) is y = 6x + 6.
To find the equation of the tangent line to the curve at the given point, we need to first find the slope of the tangent line.
The slope of the tangent line can be found by taking the derivative of the given function with respect to x.
Let's start by finding the derivative of y with respect to x.
dy/dx = d/dx (6e^x cos x)
To find the derivative of the function, we can use the product rule. Let's differentiate each term separately.
Let u = 6e^x, and v = cos x.
Using the product rule:
dy/dx = du/dx * v + u * dv/dx
Now let's find du/dx and dv/dx.
Taking the derivative of u = 6e^x:
du/dx = 6 * d/dx (e^x)
Since the derivative of e^x is e^x, we have:
du/dx = 6 * e^x
Taking the derivative of v = cos x:
dv/dx = d/dx (cos x)
The derivative of cos x is -sin x:
dv/dx = -sin x
Now, substitute the values back into the equation:
dy/dx = du/dx * v + u * dv/dx
= (6 * e^x) * cos x + (6e^x) * (-sin x)
Simplifying the equation further:
dy/dx = 6e^x * cos x - 6e^x * sin x
Now that we have the slope of the tangent line, we can find the equation using the point-slope form.
The point-slope form of a line is given by: y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Given the point (0, 6), the equation becomes:
y - 6 = (6e^x * cos x - 6e^x * sin x)(x - 0)
Simplifying:
y - 6 = (6e^x * cos x - 6e^x * sin x)x
Finally, rearranging the equation to the slope-intercept form:
y = (6e^x * cos x - 6e^x * sin x)x + 6
Thus, the equation of the tangent line to the curve at the point (0, 6) is y = (6e^x * cos x - 6e^x * sin x)x + 6.
y' = 6e^x(cosx - sinx)
y'(0) = 6(1-0) = 6
so, you have a point and a slope. The rest is just algebra I.