a) determine the equation of the line that is perpendicular to the tangent to y=5x^2 at (1,5).
b) determine the equation of the line that passes through (2,2) and is parallel to the line tangent to y=-3x^3-2x at (-1,5).
(a)
y=5x²
dy/dx=5*2x=10x, slope of tangent
slope of line perpendicular to tangent
= -1/(10x)
At (1,5), slope of perpendicular line
= -1/(10*1) = -1/10
Line passing through (1,5)
(y-5)=-(1/10)(x-1)
y =-(1/10)x + 5.10
Test for perpendicularity at (1,5)
-(1/10) * (10*1)
= -1 Line and tangent are perpendicular.
(b) y=-3x^3-2x at (-1,5).
dy/dx = -9x² -2 = -9-2 = -11 at (-1,5)
Slope of tangent = -11
Line passing through (2,2) with slope of -11
(y-2) = -11(x-2)
y-2 = -11x + 22
y = -11x + 24
slope of -8y-3x+5=10x
a) To find the equation of the line that is perpendicular to the tangent to y = 5x^2 at (1,5), we need to find the slope of the tangent line first.
1. Find the derivative of y with respect to x: y' = d/dx(5x^2) = 10x.
2. Substitute x = 1 into the derivative to find the slope of the tangent line: m = 10(1) = 10.
Since the tangent line has a slope of 10, the perpendicular line will have a negative reciprocal slope. Thus, the slope of the perpendicular line is -1/10.
3. Use the point-slope form of a line to find the equation of the line passing through (1,5) with a slope of -1/10:
y - y1 = m(x - x1)
y - 5 = (-1/10)(x - 1)
y - 5 = (-1/10)x + 1/10
y = (-1/10)x + 51/10
Therefore, the equation of the line perpendicular to the tangent to y = 5x^2 at (1,5) is y = (-1/10)x + 51/10.
b) To find the equation of the line that passes through (2,2) and is parallel to the line tangent to y = -3x^3 - 2x at (-1,5), we again need to find the slope of the tangent line.
1. Find the derivative of y with respect to x: y' = d/dx(-3x^3 - 2x) = -9x^2 - 2.
2. Substitute x = -1 into the derivative to find the slope of the tangent line: m = -9(-1)^2 - 2 = -9 - 2 = -11.
Since the tangent line has a slope of -11, the parallel line will also have a slope of -11.
3. Use the point-slope form of a line to find the equation of the line passing through (2,2) with a slope of -11:
y - y1 = m(x - x1)
y - 2 = (-11)(x - 2)
y - 2 = -11x + 22
y = -11x + 24
Therefore, the equation of the line that passes through (2,2) and is parallel to the tangent line of y = -3x^3 - 2x at (-1,5) is y = -11x + 24.
a) To find the equation of a line that is perpendicular to the tangent to y = 5x^2 at (1,5), we need to do the following steps:
Step 1: Find the slope of the tangent line.
The slope of a tangent line to a curve at a given point is equal to the derivative of the curve at that point. In this case, the derivative of y = 5x^2 is dy/dx = 10x. Evaluating the derivative at x = 1, we get dy/dx = 10(1) = 10. Therefore, the slope of the tangent line is 10.
Step 2: Find the slope of the line perpendicular to the tangent.
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line. So, the slope of the line perpendicular to the tangent line is -1/10.
Step 3: Use the slope-intercept form of the line to determine the equation of the line.
The slope-intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept. Since we know the slope is -1/10 and the line passes through the point (1,5), we can substitute these values into the equation to solve for b:
5 = (-1/10)(1) + b
5 = -1/10 + b
b = 5 + 1/10
b = 51/10
Therefore, the equation of the line that is perpendicular to the tangent to y = 5x^2 at (1,5) is y = (-1/10)x + 51/10.
b) To determine the equation of the line that passes through (2,2) and is parallel to the line tangent to y = -3x^3 - 2x at (-1,5), we need to follow these steps:
Step 1: Find the slope of the tangent line.
Just like in part (a), the slope of the tangent line is equal to the derivative of the curve at the given point. The derivative of y = -3x^3 - 2x is dy/dx = -9x^2 - 2. Evaluating the derivative at x = -1, we get dy/dx = -9(-1)^2 - 2 = -9 - 2 = -11. Therefore, the slope of the tangent line is -11.
Step 2: Find the slope of the line parallel to the tangent.
A line parallel to another line will have the same slope. So, the slope of the line parallel to the tangent line is also -11.
Step 3: Use the slope-intercept form of the line to determine the equation of the line.
Using the point-slope form of a line, which is y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is a point it passes through, we can substitute the values of the slope (-11) and the point (2,2) into the equation:
y - 2 = -11(x - 2)
Simplifying the equation, we get:
y - 2 = -11x + 22
Rearranging the equation in slope-intercept form, we have:
y = -11x + 24
Therefore, the equation of the line that passes through (2,2) and is parallel to the line tangent to y = -3x^3 - 2x at (-1,5) is y = -11x + 24.