Find all points of the graph of y=7x^2+7x whose tangent lines are perpendicular to the line -49y=x
y = -1/49 x
the slope of the perpendicular is
... -(-1/49) = 49
the 1st derivative is the slope of the tangent ... dy/dx = 14x + 7
14x + 7 = 49
find x, then find y from the original function
-49y=x has slope -1/49
so, you want lines with slope 49
since y' = 7(2x+1), that means x=3.
y(3) = 84
so, the line is
y-84 = 49(x-3)
See
http://www.wolframalpha.com/input/?i=plot+y%3D7x%5E2%2B7x,+y%3D49(x-3)%2B84+for+x%3D0..5
To find the points on the graph of y = 7x^2 + 7x where the tangent lines are perpendicular to the line -49y = x, we need to find the slope of the line -49y = x and then find the slopes of the tangent lines to the curve y = 7x^2 + 7x.
Let's start by finding the slope of the line -49y = x:
Rearrange the equation to get y = -(1/49)x:
Comparing it with the slope-intercept form y = mx + b, we can see that the slope m is -(1/49).
Since the slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line, the slope of the tangent lines to the curve y = 7x^2 + 7x should be the negative reciprocal of -(1/49). Let's calculate it:
The negative reciprocal of -(1/49) is 49. So, the slope of the tangent lines should be 49.
Now, let's find the points where the tangent lines have a slope of 49 on the curve y = 7x^2 + 7x:
Differentiate the equation y = 7x^2 + 7x to find the derivative:
dy/dx = 14x + 7
Setting 14x + 7 equal to 49 (the required slope of the tangent lines), we get:
14x + 7 = 49
Subtracting 7 from both sides:
14x = 42
Dividing both sides by 14:
x = 3
Now, we have the x-coordinate of the point(s) where the tangent lines have a slope of 49. To find the corresponding y-coordinate, substitute x = 3 into the original equation y = 7x^2 + 7x:
y = 7(3^2) + 7(3)
y = 63
Therefore, the point on the graph of y = 7x^2 + 7x where the tangent lines are perpendicular to the line -49y = x is (3, 63).
So, there is only one point on the graph of y = 7x^2 + 7x where the tangent lines are perpendicular to the line -49y = x, and it is (3, 63).
To find the points on the graph of y = 7x^2 + 7x whose tangent lines are perpendicular to the line -49y = x, we need to find the slope of the line -49y = x and then find the slopes of the tangent lines of the given graph.
Step 1: Find the slope of the line -49y = x:
The equation -49y = x can be rewritten in slope-intercept form (y = mx + b) as follows:
-49y = x
Divide both sides by -49:
y = x/-49
Comparing this equation with y = mx + b, we can see that the slope of the line is -1/49.
Step 2: Find the slope of the tangent lines of the graph y = 7x^2 + 7x:
The derivative of y = 7x^2 + 7x can be found by taking the derivative of each term separately. Applying the power rule, we get:
d/dx (7x^2) = 14x
d/dx (7x) = 7
So, the derivative of y = 7x^2 + 7x is dy/dx = 14x + 7. This represents the slope of the tangent line at any point on the graph.
Step 3: Find the x-values where the slopes of the tangent lines are perpendicular:
Since two lines are perpendicular if and only if their slopes are negative reciprocals of each other, we need to find the values of x for which the slope, dy/dx = 14x + 7, is equal to the negative reciprocal of the slope of the line -1/49.
The negative reciprocal of -1/49 is 49/1 = 49.
So we have the equation:
14x + 7 = 49
Subtract 7 from both sides:
14x = 42
Divide both sides by 14:
x = 3
Therefore, the x-coordinate of the points where the tangent lines are perpendicular is x = 3.
Step 4: Find the corresponding y-values:
To find the corresponding y-values, substitute the x-value (x = 3) into the equation y = 7x^2 + 7x:
y = 7(3)^2 + 7(3)
y = 63 + 21
y = 84
So, the points on the graph of y = 7x^2 + 7x whose tangent lines are perpendicular to the line -49y = x are (3, 84).