By finding the slopes of the tangent lines to the curve of y=(1/3)x^3+5x at the points where x=3 and x=6, find the acute angle between these lines at the point where they cross.
I don't have any idea where to start. An explanation would be appreciated.
Please and thanks.
slope = dy/dx = x^2 + 5
if x = 3
y = 9+15 = 24 so through (3,24)
here slope = 9+5 = 14
y = mx+b = 24 = 14(3) +b so b = -18
so
y = 14 x - 18
if x = 6
slope = 36+5 = 41
y = 72 + 30 = 102
102 = 41*6 + b
b = -144
y = 41 x - 144
so find intersection of lines
y = 41 x -144 and
y = 14 x - 18
Then you know how to find the slope of each line at that point and find the angle between
tan^-1(slope 1 - slope 2)
Not tan^-1(m1 - m2)
but rather
tan^-1 m1 - tan^-1 m2
whoops, yes
To find the slopes of the tangent lines at the points where x=3 and x=6, we need to first find the derivative of the given function. The derivative gives us the slope of the tangent line at any point on the curve.
The given function is y=(1/3)x^3+5x. To find its derivative, we use the power rule for differentiation. The power rule states that for a term of the form ax^n, the derivative is given by d/dx(ax^n) = anx^(n-1).
Applying the power rule to each term in the function, we differentiate (1/3)x^3 to get (1/3)(3)x^(3-1), which simplifies to x^2. Similarly, we differentiate 5x to get 5.
Therefore, the derivative of y=(1/3)x^3+5x is dy/dx = x^2 + 5.
Now that we have the derivative, we can find the slopes of the tangent lines at x=3 and x=6.
When x=3, we substitute this value into the derivative:
dy/dx = (3)^2 + 5 = 9 + 5 = 14.
So, the slope of the tangent line at x=3 is 14.
Similarly, when x=6, we substitute this value into the derivative:
dy/dx = (6)^2 + 5 = 36 + 5 = 41.
So, the slope of the tangent line at x=6 is 41.
Now, we know the slopes of the tangent lines at the points where x=3 and x=6. To find the acute angle between these lines at the point where they cross, we can use the formula for the angle between two lines, given by tan(theta) = |(m1 - m2) / (1 + m1*m2)|, where m1 and m2 are the slopes of the lines.
Substituting the slopes we found earlier, we have tan(theta) = |(14 - 41) / (1 + 14*41)| = |-27 / (1 + 574)|.
To find the acute angle theta, we take the inverse tangent (arctan) of the right-hand side of the equation:
theta = arctan(-27 / (1 + 574)).
Evaluating this using a calculator, we find that the acute angle between the tangent lines is approximately 1.625 radians or 93.20 degrees.