Find the equation of the tangent line to the graph of (x+y)^3=27x at the point (1,2)
take the differential
3(x+y)^2 (1+y')=27
y'+1=9/(x+y)^2
y'=9/(x+y)^2 -1 = m
this is the slope
y=mx+b at the point 1,2
solve for b
when x=1, y=2
then the equation is
y=mx + b
I understand , but when you plug in the x and y in and I get y' =0 so would that be my slope?
To find the equation of the tangent line to the graph of the equation (x+y)^3 = 27x, we can follow these steps:
Step 1: Rewrite the equation in a more manageable form.
Start by expanding the left side of the equation using the binomial theorem:
x^3 + 3x^2y + 3xy^2 + y^3 = 27x
Next, rearrange the terms to isolate y:
3x^2y + 3xy^2 + y^3 = 27x - x^3
Since we're looking for the tangent line at a specific point (1,2), we can substitute these values to get:
3(1)^2(2) + 3(1)(2)^2 + y^3 = 27(1) - (1)^3
6 + 12 + y^3 = 27 - 1
18 + y^3 = 26
y^3 = 8
Taking the cube root of both sides:
y = 2
So, at the point (1,2), the value of y is 2.
Step 2: Find the derivative of the equation.
The derivative of both sides of the equation will give us the slope of the tangent line at any given point. So, let's differentiate the equation (x+y)^3 = 27x.
To differentiate (x+y)^3, we will use the chain rule. Let's first differentiate with respect to x:
d/dx[(x+y)^3] = d/dx[27x]
3(x+y)^2 * (1+y') = 27
Step 3: Find the slope at the point (1,2).
We need to find the slope of the tangent line, which corresponds to the derivative when x = 1 and y = 2.
Substituting these values into the derived equation:
3(1+2)^2 * (1+y') = 27
3(9)(1+y') = 27
27(1+y') = 27
1+y' = 1
y' = 1-1
y' = 0
So, the slope of the tangent line at the point (1,2) is 0.
Step 4: Write the equation of the tangent line.
Now we have the slope of the tangent line (m = 0) and the coordinates of the point (1,2).
Using the point-slope form of a linear equation, we can write the equation of the tangent line as:
y - y1 = m(x - x1), where (x1, y1) is the given point.
Plugging in the values:
y - 2 = 0(x - 1)
y - 2 = 0
y = 2
The equation of the tangent line to the graph of (x+y)^3 = 27x at the point (1,2) is y = 2.