the normal line to the curve y=f(x)=2x^2 at a point in the first quadrant passes through the point (0,3/4). find an equation of this normal line
take a minute to notice that the point is on the axis of symmetry...
To find the equation of the normal line to the curve y = f(x) = 2x^2 at a point in the first quadrant, we need to find the slope of the tangent line and then determine the negative reciprocal of that slope to get the slope of the normal line. Let's proceed step by step.
Step 1: Find the derivative of f(x) to get the slope of the tangent line at any point on the curve.
f(x) = 2x^2
Taking the derivative using the power rule, we have:
f'(x) = 4x
Step 2: Substitute the x-coordinate of the point of interest into the derivative formula to find the slope of the tangent line at that point.
Let's say the x-coordinate of the point of interest is x1.
The slope of the tangent line, m, is given by:
m = f'(x1) = 4x1
Step 3: Determine the negative reciprocal of the slope to find the slope of the normal line.
The slope of the normal line, m_n, is given by:
m_n = -1/m
Step 4: Use the slope-intercept form of a line, y = mx + b, and the given point (0, 3/4) to find the y-intercept, b.
Substituting the point (0, 3/4) and the slope m_n into the slope-intercept form, we have:
3/4 = m_n * 0 + b
3/4 = b
Step 5: Write the equation of the normal line using the slope-intercept form.
The equation of the normal line is:
y = m_n * x + b, where m_n is the negative reciprocal of the slope and b is the y-intercept we found.
Substituting the values we found, the equation becomes:
y = (-1/m) * x + 3/4
Thus, the equation of the normal line to the curve y = f(x) = 2x^2, that passes through the point (0, 3/4), is:
y = (-1/4x1) * x + 3/4
To find the equation of the normal line to the curve at a given point, we first need to determine the slope of the normal line, which is the negative reciprocal of the slope of the tangent line at that point. Here's how you can do that:
Step 1: Find the derivative of the function y = f(x) = 2x^2.
To find the derivative, use the power rule, which states that if y = ax^n, then dy/dx = nax^(n-1).
For y = 2x^2, the derivative dy/dx is obtained by applying the power rule, giving: dy/dx = 2 * 2x^(2-1) = 4x.
Step 2: Determine the slope of the tangent line.
Plug the x-coordinate of the given point into the derivative we found in step 1. Since the slope of the tangent line is given by the derivative at the point, we have:
Slope of the tangent line = 4x.
Using the x-coordinate of the given point, substitute the value into the slope equation:
Slope of the tangent line at the point (x, y) = 4x.
Step 3: Find the slope of the normal line.
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Take the reciprocal of the tangent line slope, then negate it (multiply by -1):
Slope of the normal line = -1 / (4x).
Step 4: Use the given point to find the y-intercept of the normal line.
Substitute the coordinates (0, 3/4) into the equation y = mx + b, where m is the slope of the normal line and b is the y-intercept we want to find:
3/4 = (-1 / (4x)) * 0 + b
3/4 = b
b = 3/4
Step 5: Write the equation of the normal line.
Now that we have the slope (-1 / (4x)) and the y-intercept (3/4), we can write the equation of the normal line in the form y = mx + b using the point-slope form:
y = (-1 / (4x)) * x + 3/4
Simplifying further, we get:
y = -1 / 4x + 3/4
Therefore, the equation of the normal line to y = f(x) = 2x^2 at the point (0, 3/4) in the first quadrant is y = -1 / 4x + 3/4.
Sure, let's find the equation of this normal line through the point (0,3/4). Since we know that the line is normal to the curve, we can use the fact that the slope of the normal line is the negative reciprocal of the slope of the curve at the given point.
The derivative of y = f(x) = 2x^2 is y' = 4x. Let's find the slope of the curve at the point (x, y).
At the point (x, y), the slope of the curve is y' = 4x. Since the normal line is perpendicular to the slope of the curve, the slope of the normal line will be -1/4x.
Since the normal line passes through the point (0,3/4), we can use the point-slope form of a linear equation to find the equation of the normal line:
y - y1 = m(x - x1)
Substituting the coordinates (0,3/4) and slope -1/4x, we get:
y - 3/4 = -(1/4x)(x - 0)
Simplifying this equation, we have:
y - 3/4 = -1/4x
Rearranging the equation to the standard form Ax + By = C, we get:
1/4x + y = 3/4
Multiplying both sides of the equation by 4 to get rid of the fractions, we have:
x + 4y = 3
So, the equation of the normal line to the curve y = f(x) = 2x^2 at a point in the first quadrant that passes through the point (0,3/4) is x + 4y = 3.
Hope that "lines" up with what you were looking for!