The coordinates of the point where the normal to the curve y= 1/3 x^3 + 1/2 x^2 + x at x=1 intersects the y-axis are what?
dy/dx = x^2 + x + 1
at x=1 ,slope of tangent is 1+1+1 = 3
so slope of normal is -1/3
when x=1, y = 1/3 + 1/2 + 1 = 11/6
so equation of normal:
y = (-1/3)x + b, at (1, 11/6)
11/6 = -1/3 + b
b = 3/2
y = (-1/3)x + 13/6
for y-intercept, let x = 0
y = 13/6
it cuts at (0,13/6)
check my arithmetic
To find the coordinates of the point where the normal to the curve intersects the y-axis, we need to follow these steps:
Step 1: Find the derivative of the given curve to find the slope of the tangent line at x=1.
Step 2: Find the negative reciprocal of the slope of the tangent line to get the slope of the normal line.
Step 3: Use the point-slope form of a line to find the equation of the normal line passing through the point (1, y(1)).
Step 4: Find the y-intercept of the normal line to get the point where it intersects the y-axis.
Let's calculate step-by-step:
Step 1:
Taking the derivative of the curve y = (1/3)x^3 + (1/2)x^2 + x:
dy/dx = d/dx[(1/3)x^3 + (1/2)x^2 + x]
Using the power rule of differentiation, we have:
dy/dx = (1/3)(3x^2) + (1/2)(2x) + 1
dy/dx = x^2 + x + 1
At x = 1, the slope of the tangent line is given by:
m = dy/dx = 1^2 + 1 + 1 = 3
Step 2:
The slope of the normal line is the negative reciprocal of the slope of the tangent line:
m_normal = -1/m = -1/3
Step 3:
Using the point-slope form of a line, the equation of the normal line passing through the point (1, y(1)) is given by:
y - y1 = m_normal(x - x1)
Substituting the values of x1 = 1 and y1 = y(1), we get:
y - y(1) = (-1/3)(x - 1)
Step 4:
To find the point where the normal line intersects the y-axis, we need to set x = 0 in the equation found in step 3.
Setting x = 0, we have:
y - y(1) = (-1/3)(0 - 1)
y - y(1) = (-1/3)(-1)
y - y(1) = 1/3
Now, solving for y, we get:
y = y(1) + 1/3
Therefore, the coordinates of the point where the normal to the curve y = (1/3)x^3 + (1/2)x^2 + x at x=1 intersects the y-axis are (0, y(1) + 1/3).
To find the coordinates of the point where the normal to the curve intersects the y-axis, we need to follow a few steps:
Step 1: Find the derivative of the curve equation.
The derivative will give us the slope of the tangent line at any given point on the curve. In this case, we need the slope of the normal line, so we'll find the derivative of y with respect to x.
The derivative of y = (1/3)x^3 + (1/2)x^2 + x is:
dy/dx = d/dx[(1/3)x^3 + (1/2)x^2 + x]
= x^2 + x + 1
Step 2: Find the slope of the normal line.
The slope of the normal line is the negative reciprocal of the slope of the tangent line. So, we need to find the negative reciprocal of the derivative we found in step 1.
The slope of the normal line = -1 / (x^2 + x + 1)
Step 3: Find the equation of the normal line.
To find the equation of the normal line, we need both the slope and a point on the line. We are given that the normal line passes through the point (x, y), where x = 1 and y is the y-coordinate of the point on the curve at x = 1.
To find the y-coordinate of the point on the curve at x = 1, substitute x = 1 into the original curve equation:
y = (1/3)(1)^3 + (1/2)(1)^2 + 1
= 1/3 + 1/2 + 1
= 1 + 3/6 + 6/6
= 10/6
= 5/3
So, the normal line passes through the point (1, 5/3).
Step 4: Write the equation of the normal line using the point-slope form.
The equation of a line in point-slope form is given by: y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.
Plugging in the values, we have:
y - (5/3) = (-1 / (1^2 + 1 + 1))(x - 1)
Simplifying:
y - (5/3) = (-1 / 3)(x - 1)
3y - 5 = -x + 1
3y = -x + 6
Step 5: Find the point where the normal line intersects the y-axis.
To find the point where the line intersects the y-axis, we set x = 0 in the equation we derived in step 4.
3y = -0 + 6
3y = 6
y = 2
Therefore, the coordinates of the point where the normal line intersects the y-axis are (0, 2).