Determine the equation of the tangent and normal to the graph x(xy+y^3)-24=0 at the point (1,2)
The point (1,2) is not on the graph!
Apologies for the mistake. Let's correct it.
To determine the equation of the tangent and normal at the point (1,2), we need to find the derivatives and evaluate them at that point.
Given the equation x(xy + y^3) - 24 = 0, let's find the derivatives with respect to x:
d/dx (x(xy + y^3) - 24) = 0
Using the product rule, we have:
d/dx (x^2y + xy^3 - 24) = 0
Taking the derivatives of each term separately, we get:
2xy + x^2(dy/dx) + y^3 + 3xy^2(dy/dx) = 0
Now, let's substitute x = 1 and y = 2 into this equation:
2(1)(2) + (1)^2(dy/dx) + (2)^3 + 3(1)(2)^2(dy/dx) = 0
4 + dy/dx + 8 + 12(dy/dx) = 0
13(dy/dx) + dy/dx = -12
14(dy/dx) = -12
dy/dx = -12/14
Simplifying further, we get:
dy/dx = -6/7
Thus, the slope of the tangent line at the point (1,2) is -6/7.
The slope of the normal line is the negative reciprocal of the slope of the tangent line. So, the slope of the normal line is:
m_normal = -1/(dy/dx) = -1/(-6/7) = 7/6.
Now that we have the slope of the tangent line and the slope of the normal line, we can find their equations using the point-slope form.
For the tangent line:
y - 2 = (-6/7)(x - 1)
Simplifying:
7y - 14 = -6x + 6
Rearranging, we get:
6x + 7y = 20
This is the equation of the tangent line to the graph at the point (1,2).
For the normal line:
y - 2 = (7/6)(x - 1)
Simplifying:
6y - 12 = 7x - 7
Rearranging, we get:
7x - 6y = 5
This is the equation of the normal line to the graph at the point (1,2).
To determine the equation of the tangent and normal to the graph at the point (1,2), we need to find the slope of the tangent line and the slope of the normal line at that point.
Step 1: Find the derivative of the equation with respect to x.
Let's differentiate the given equation, x(xy + y^3) - 24 = 0, with respect to x.
Differentiating both sides of the equation:
d/dx(x(xy + y^3) - 24) = d/dx(0)
Using the product rule and the chain rule, we get:
x(dy/dx)(xy + y^3) + (xy + y^3) + x(2y + 3y^2)(dy/dx) = 0
Step 2: Plug in the point (1,2) to find the slope at that point.
Substitute x=1 and y=2 into the equation derived in Step 1.
1(2(dy/dx)(2 + 2^3) + (2 + 2^3) + 1(2(2) + 3(2)^2)(dy/dx) = 0
Simplifying the equation gives:
4(dy/dx) + 30(dy/dx) = -22
Combining like terms:
34(dy/dx) = -22
Step 3: Solve for dy/dx, which gives the slope of the tangent line.
Divide both sides of the equation by 34:
(dy/dx) = -22/34
Simplifying the equation:
(dy/dx) = -11/17
So, the slope of the tangent line at the point (1,2) is -11/17.
Step 4: Determine the slope of the normal line.
The slope of the normal line is the negative reciprocal of the slope of the tangent line.
Therefore, the slope of the normal line will be:
m = -1 / (dy/dx) = -1 / (-11/17) = 17/11
So, the slope of the normal line is 17/11.
Step 5: Use the point-slope form to find the equations of the tangent and normal lines.
Using the point-slope form, we can find the equation of the tangent line and the normal line.
The point-slope form of a line is given by:
y - y1 = m(x - x1)
For the tangent line:
Substituting the values of x1 = 1, y1 = 2, and m = -11/17 into the equation, we get:
y - 2 = (-11/17)(x - 1)
Simplifying the equation gives:
17y - 34 = -11x + 11
Rearranging the terms:
11x + 17y = 45
So, the equation of the tangent line to the graph at the point (1,2) is 11x + 17y = 45.
For the normal line:
Substituting the values of x1 = 1, y1 = 2, and m = 17/11 into the equation, we get:
y - 2 = (17/11)(x - 1)
Simplifying the equation gives:
11y - 22 = 17x - 17
Rearranging the terms:
17x - 11y = 5
So, the equation of the normal line to the graph at the point (1,2) is 17x - 11y = 5.
Therefore, the equation of the tangent line is 11x + 17y = 45 and the equation of the normal line is 17x - 11y = 5.
To find the equation of the tangent and normal, we need to find the slope of the tangent line and the slope of the normal line at the given point.
First, let's find the slope of the tangent line. We can do this by taking the derivative of the equation with respect to x:
d/dx (x(xy+y^3)-24) = 0
Using the product rule, we have:
d/dx(x^2y + xy^3 - 24) = 0
Differentiating each term separately, we get:
2xy + x^2(dy/dx) + y^3 + 3xy^2(dy/dx) = 0
Now, let's plug in the coordinates of the given point (1,2) into this equation:
2(1)(2) + (1)^2(dy/dx) + (2)^3 + 3(1)(2)^2(dy/dx) = 0
4 + dy/dx + 8 + 12(dy/dx) = 0
13(dy/dx) + dy/dx = -12
14(dy/dx) = -12
dy/dx = -12/14
Simplifying, we get:
dy/dx = -6/7
So, the slope of the tangent line at the point (1,2) is -6/7.
Next, let's 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, the slope of the normal line is:
m_normal = -1/(dy/dx) = -1/(-6/7) = 7/6
Now we have the slope of the tangent line and the slope of the normal line. To find the equation of a line, we need the slope and a point on the line. We already have the point (1,2).
Using the point-slope form of a line, the equation of the tangent line is:
y - 2 = (-6/7)(x - 1)
Simplifying, we get:
7y - 14 = -6x + 6
Re-arranging, we have:
6x + 7y = 20
This is the equation of the tangent line to the graph at the point (1,2).
To find the equation of the normal line, we use the point-slope form again:
y - 2 = (7/6)(x - 1)
Simplifying, we get:
6y - 12 = 7x - 7
Re-arranging, we have:
7x - 6y - 5 = 0
This is the equation of the normal line to the graph at the point (1,2).