To find the equation of the tangent line at (1,2), we need to find the derivative of the curve at that point, which will give us the slope of the tangent line.
Taking the derivative of the curve with respect to x, we get:
2xy + x²dy/dx + 3y²dy/dx = 1
Simplifying and solving for dy/dx, we get:
dy/dx = (1 - 2xy) / (3y² - x²)
At point (1,2),
x = 1, y = 2
So,
dy/dx = (1 - 2(1)(2)) / (3(2²) - 1²)
dy/dx = -3/11
Therefore, the slope of the tangent line at (1,2) is -3/11.
Using the point-slope equation of a line, we can write the equation of the tangent line as:
y - 2 = (-3/11)(x - 1)
Simplifying, we get:
y = (-3/11)x + 23/11
Therefore, the equation of the tangent line to the curve x²y + y³ = x - 7 at point (1,2) is y = (-3/11)x + 23/11.
find the equation of the tangent to the curve x²y+y³=x-7 at (1,2)
5 answers
wrong again, Bot!
2xy + x²y' + 3y²y' = 1
y' = (1-2xy)/(x²+3y²)
Now the next error is that the point (1,2) is not on the graph!
x²y+y³=x-7
contains the point (1,-1.63) and no point where y=2.
2xy + x²y' + 3y²y' = 1
y' = (1-2xy)/(x²+3y²)
Now the next error is that the point (1,2) is not on the graph!
x²y+y³=x-7
contains the point (1,-1.63) and no point where y=2.
I apologize for my mistakes in the previous response. Thank you for correcting me.
Taking the derivative of the curve with respect to x, we get:
2xy + x²dy/dx + 3y²dy/dx = 1
Simplifying and solving for dy/dx, we get:
dy/dx = (1 - 2xy) / (x² + 3y²)
At point (1,-1.63),
x = 1, y = -1.63
So,
dy/dx = (1 - 2(1)(-1.63)) / (1² + 3(-1.63)²)
dy/dx = 0.499
Therefore, the slope of the tangent line at (1,-1.63) is 0.499.
Using the point-slope equation of a line, we can write the equation of the tangent line as:
y - (-1.63) = (0.499)(x - 1)
Simplifying, we get:
y = 0.499x - 2.129
Therefore, the equation of the tangent line to the curve x²y + y³ = x - 7 at point (1,-1.63) is y = 0.499x - 2.129.
Taking the derivative of the curve with respect to x, we get:
2xy + x²dy/dx + 3y²dy/dx = 1
Simplifying and solving for dy/dx, we get:
dy/dx = (1 - 2xy) / (x² + 3y²)
At point (1,-1.63),
x = 1, y = -1.63
So,
dy/dx = (1 - 2(1)(-1.63)) / (1² + 3(-1.63)²)
dy/dx = 0.499
Therefore, the slope of the tangent line at (1,-1.63) is 0.499.
Using the point-slope equation of a line, we can write the equation of the tangent line as:
y - (-1.63) = (0.499)(x - 1)
Simplifying, we get:
y = 0.499x - 2.129
Therefore, the equation of the tangent line to the curve x²y + y³ = x - 7 at point (1,-1.63) is y = 0.499x - 2.129.
The first step in these type of tangent questions is to check if the given
point does actually lie on the graph.
the bot did not do that, and the point is not on the graph, so any
calculations that follow that would be bogus.
Of course oobleck picked up on that point, again showing the
superiority of a thinking human mind over the AI.
point does actually lie on the graph.
the bot did not do that, and the point is not on the graph, so any
calculations that follow that would be bogus.
Of course oobleck picked up on that point, again showing the
superiority of a thinking human mind over the AI.
You are absolutely right. Checking whether the given point lies on the graph is an important step in finding the equation of the tangent line. I apologize for the mistake and thank you and oobleck for bringing it to my attention. As an AI language model, my responses are generated based on patterns in large datasets, and I am constantly learning and improving based on feedback from users like you.