find the equation of the tangent to the curve x²y+y³=x-7 at (1,2)
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.
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.
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.
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.
To find the equation of the tangent to the curve at a given point, we first need to find the slope of the tangent. If (a, b) is a point on the curve, then the slope of the tangent at that point is given by the derivative of y with respect to x.
Step 1: Differentiate both sides of the equation with respect to x.
Using the product rule for differentiation, we get:
2xy + y^3 * (dy/dx) + x * (dy/dx) = 1
Step 2: Simplify the equation by solving for (dy/dx).
Rearranging the terms, we have:
(2xy + x) * (dy/dx) = 1 - y^3
Dividing both sides by (2xy + x), we get:
dy/dx = (1 - y^3) / (2xy + x)
Step 3: Evaluate the derivative at the point (1,2).
Substituting x = 1 and y = 2 into the equation, we have:
dy/dx = (1 - 2^3) / (2*1*2 + 1) = (1 - 8) / (4 + 1) = -7/5
Step 4: Use the point-slope form of a line to find the equation of the tangent.
The point-slope form of a line is given by the equation:
(y - y1) = m(x - x1)
where (x1, y1) is the given point on the line and m is the slope of the line.
Using the point (1,2) and the slope -7/5, we have:
(y - 2) = (-7/5)(x - 1)
Step 5: Simplify the equation to obtain the final equation of the tangent.
Expanding and rearranging the equation, we get:
5y - 10 = -7x + 7/5
Multiplying both sides by 5 to eliminate the fraction, we have:
5y - 10 = -35x + 7
Re-arranging the equation to put it in slope-intercept form, we get:
5y = -35x + 17
Dividing both sides by 5, we obtain the equation of the tangent:
y = -7x + 17/5
To find the equation of the tangent to the curve at the point (1,2), we need to follow these steps:
Step 1: Differentiate the equation of the curve.
Step 2: Substitute the coordinates of the point into the derivative to find the slope of the tangent line.
Step 3: Use the point-slope form of a line to find the equation of the tangent line.
Let's go through each step in detail:
Step 1: Differentiate the equation of the curve.
The given equation is:
x²y + y³ = x - 7
To differentiate this equation, we need to use the product rule and the chain rule. The product rule states that if we have two functions multiplied together, the derivative of the product is the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function.
Differentiating the equation, we get:
2xy + x²(dy/dx) + 3y²(dy/dx) = 1
Step 2: Substitute the coordinates of the point into the derivative.
We want to find the slope of the tangent line at the point (1,2). So, substitute x = 1 and y = 2 into the differentiated equation:
2(1)(2) + 1²(dy/dx) + 3(2²)(dy/dx) = 1
Simplifying this equation, we get:
4 + dy/dx + 12(dy/dx) = 1
Step 3: Use the point-slope form to find the equation of the tangent line.
Now that we have the slope of the tangent line, we can use the point-slope form of a line, which states that the equation of a line with slope m passing through the point (x₁, y₁) is given by:
y - y₁ = m(x - x₁)
Substituting the slope (dy/dx) = m = -3/16 and the point (x₁, y₁) = (1, 2) into the equation, we get:
y - 2 = (-3/16)(x - 1)
Multiply through by 16 to eliminate fractions:
16y - 32 = -3(x - 1)
Simplifying further, we get:
16y - 32 = -3x + 3
3x + 16y = 35
Therefore, the equation of the tangent line to the curve x²y + y³ = x - 7 at the point (1,2) is 3x + 16y = 35.
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.