Find a unit vector that is parallel to the line tangent to the parabola y = x^2 at the point (4, 16).
the slope at (4,16) is 8.
So, the tangent vector is
i + 8j
The unit vector is just that, divided by the length: √65
To find a unit vector that is parallel to the line tangent to the parabola y = x^2 at the point (4, 16), we need to determine the slope of the tangent line at that point.
The slope of a tangent line to a function at a specific point can be found using the derivative.
Taking the derivative of the equation y = x^2, we get dy/dx = 2x.
Now, let's find the slope of the tangent line at the point (4, 16):
dy/dx = 2(4) = 8.
Therefore, the slope of the tangent line at the point (4, 16) is 8.
Now, a unit vector in the direction of a line with slope m can be obtained by dividing the vector (1, m) by its magnitude.
So, the direction vector of the tangent line is (1, 8).
To obtain a unit vector, divide this vector by its magnitude as follows:
Magnitude of (1, 8) = sqrt(1^2 + 8^2) = sqrt(65).
Finally, dividing the vector (1, 8) by its magnitude, we get:
(1/sqrt(65), 8/sqrt(65))
Hence, a unit vector that is parallel to the line tangent to the parabola y = x^2 at the point (4, 16) is (1/sqrt(65), 8/sqrt(65)).
To find a unit vector parallel to the line tangent to the parabola y = x^2 at the point (4, 16), we can follow these steps:
Step 1: Find the derivative of the function y = x^2.
The derivative gives us the slope of the tangent line at any given point on the parabola. The derivative of y = x^2 is given by:
y' = 2x
Step 2: Evaluate the derivative at the point (4, 16).
Substitute x = 4 into the derivative to find the slope of the tangent line at that point:
y' = 2(4) = 8
Step 3: Write the equation of the tangent line at the point (4, 16).
Using the point-slope form of a line, the equation of the tangent line can be written as:
y - y₁ = m(x - x₁)
where (x₁, y₁) are the coordinates of the point (4, 16), and m is the slope of the tangent line.
Plugging in the values, we have:
y - 16 = 8(x - 4)
Step 4: Simplify the equation of the tangent line.
Distribute and simplify to get the equation in slope-intercept form (y = mx + b):
y - 16 = 8x - 32
y = 8x - 16
Step 5: Find the direction vector of the line.
The direction vector of the line is determined by the coefficients of x and y in the equation of the line. In this case, the direction vector is (8, 1).
Step 6: Convert the direction vector into a unit vector.
To convert the direction vector into a unit vector, we divide each component by the magnitude (length) of the vector.
Magnitude of the direction vector:
|d| = sqrt(8^2 + 1^2) = sqrt(65)
Unit vector:
u = (8/|d|, 1/|d|)
Simplifying,
u = (8/sqrt(65), 1/sqrt(65))
Therefore, the unit vector parallel to the line tangent to the parabola y = x^2 at the point (4, 16) is (8/sqrt(65), 1/sqrt(65)).