Find the equation in slope intercept form of the tangent line and the normal line to the curve x^2+4xy+y^2=13 at the point.(2,1).
show steps please!!!
thank you!
2x + 4x dy/dx + 4y + 2y dy/dx = 0
dy/dx(4x + 2y) = -2x - 4y
dy/dx = -2(x+2y)/( 2(2x + y) = -(x+2y)/(2x+y)
at (2,1)
dy/dx = -4/5
so slope of tangent is 4/5
slope of normal is 5/4
tangent:
y-1 = (-4/5)(x-2)
y - 1 = (-4/5) + 8/5
y = (-4/5)x + 13/5
you do the normal is the same way.
the normal line is perpendicular to the tangent line.
Since the slope of the tangent line at x is y', find y'.
2x + 4y + 4xy' + 2yy' = 0
y' = -(2x+4y)/(4x+2y) = -(x+2y)/(2x+y)
So, at (2,1) the tangent line has slope -4/5
Thus, the normal line has slope 5/4. Now you have a point and a slope, so the line is
y-1 = 5/4 (x-2)
or, in the desired form,
y = 5/4 x - 3/2
Verify here:
http://www.wolframalpha.com/input/?i=plot+x^2%2B4xy%2By^2%3D13%2C+y+%3D+5%2F4+x+-+3%2F2%2C+y+%3D+-4%2F5+x+%2B+13%2F5+for+x+%3D+-5..5
To find the equation of the tangent line and the normal line to the curve at the point (2,1), we need to first find the slope of the tangent line and then use it to find the equations of both lines. Here are the steps:
Step 1: Find the derivative of the curve equation with respect to x.
- Take the derivative of each term with respect to x.
- For the term x^2, the derivative is 2x.
- For the term 4xy, we have to use the product rule. The derivative is 4y + 4x(dy/dx).
- For the term y^2, the derivative is 2y(dy/dx).
Combining these derivatives, we get:
2x + 4y + 4x(dy/dx) + 2y(dy/dx) = 0
Step 2: Plug in the coordinates of the point (2,1) into the derivative equation to find the slope of the tangent line.
- Substitute x = 2 and y = 1 into the equation.
- Solve for dy/dx, which represents the slope of the tangent line at the given point.
Step 3: Determine 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 negative reciprocal of the slope found in Step 2.
Step 4: Use the point-slope form to find the equation of the tangent line.
- Use the point-slope form with the point (2,1) and the slope from Step 2 to find the equation of the tangent line.
Step 5: Use the point-slope form to find the equation of the normal line.
- Use the point-slope form with the point (2,1) and the slope from Step 3 to find the equation of the normal line.
Now, let's go through each step in detail.
Step 1: Find the derivative of the curve equation with respect to x.
The given equation is x^2 + 4xy + y^2 = 13.
Taking the derivative of each term with respect to x, we get:
2x + 4y + 4x(dy/dx) + 2y(dy/dx) = 0.
Step 2: Plug in the coordinates of the point (2,1) into the derivative equation to find the slope of the tangent line.
Substituting x = 2 and y = 1 into the derivative equation, we have:
2(2) + 4(1) + 8(dy/dx) + 2(1)(dy/dx) = 0.
Simplifying the equation, we get:
4 + 4 + 10(dy/dx) = 0.
8 + 10(dy/dx) = 0.
10(dy/dx) = -8.
dy/dx = -8/10.
dy/dx = -4/5.
So, the slope of the tangent line at the point (2,1) is -4/5.
Step 3: Determine the slope of the normal line.
The slope of the normal line is the negative reciprocal of the slope of the tangent line.
Taking the negative reciprocal of -4/5, we have:
Slope of the normal line = -1 / (-4/5).
Slope of the normal line = 5/4.
So, the slope of the normal line is 5/4.
Step 4: Use the point-slope form to find the equation of the tangent line.
The point-slope form of a line is given by:
y - y1 = m(x - x1),
where (x1, y1) is the point on the line and m is the slope.
Using the point (2,1) and the slope -4/5, we can write the equation of the tangent line as:
y - 1 = (-4/5)(x - 2).
Simplifying the equation, we get:
y - 1 = (-4/5)x + 8/5.
y = (-4/5)x + 8/5 + 5/5.
y = (-4/5)x + 13/5.
So, the equation of the tangent line is y = (-4/5)x + 13/5.
Step 5: Use the point-slope form to find the equation of the normal line.
Using the point (2,1) and the slope 5/4, we can write the equation of the normal line as:
y - 1 = (5/4)(x - 2).
Simplifying the equation, we get:
y - 1 = (5/4)x - 5/2.
y = (5/4)x - 5/2 + 2/2.
y = (5/4)x - 1/2.
So, the equation of the normal line is y = (5/4)x - 1/2.
Therefore, the equations in slope-intercept form of the tangent line and the normal line to the curve x^2 + 4xy + y^2 = 13 at the point (2,1) are:
Tangent line: y = (-4/5)x + 13/5.
Normal line: y = (5/4)x - 1/2.