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Answer :
To find the equation of the tangent line to the graph of the implicit function [tex]\( y^4 - 2y^3 - 5y^2 - 6x^2 - x^3 + 23 = -4x^4 \)[/tex] at the point [tex]\((1, 2)\)[/tex], follow these steps:
1. Identify the given equation: The equation of the curve is
[tex]\[
y^4 - 2y^3 - 5y^2 - 6x^2 - x^3 + 23 = -4x^4.
\][/tex]
2. Rearrange the equation: For simplicity in differentiating, write this as:
[tex]\[
y^4 - 2y^3 - 5y^2 - 6x^2 - x^3 + 23 + 4x^4 = 0.
\][/tex]
3. Calculate partial derivatives:
- To find the slope of the tangent line, you need the derivatives with respect to [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
- The partial derivative with respect to [tex]\(x\)[/tex] is:
[tex]\[
\frac{\partial}{\partial x}\left(y^4 - 2y^3 - 5y^2 - 6x^2 - x^3 + 23 + 4x^4\right) = 16x^3 - 3x^2 - 12x.
\][/tex]
- The partial derivative with respect to [tex]\(y\)[/tex] is:
[tex]\[
\frac{\partial}{\partial y}\left(y^4 - 2y^3 - 5y^2 - 6x^2 - x^3 + 23 + 4x^4\right) = 4y^3 - 6y^2 - 10y.
\][/tex]
4. Evaluate the partial derivatives at the point [tex]\((1, 2)\)[/tex]:
- Substituting [tex]\(x = 1\)[/tex] and [tex]\(y = 2\)[/tex] into the partial derivatives, we get:
- [tex]\(\left. \frac{\partial}{\partial x} \right|_{(1, 2)} = 16(1)^3 - 3(1)^2 - 12(1) = 1\)[/tex].
- [tex]\(\left. \frac{\partial}{\partial y} \right|_{(1, 2)} = 4(2)^3 - 6(2)^2 - 10(2) = -12\)[/tex].
5. Calculate the slope of the tangent line:
- The slope of the tangent line for an implicit curve is given by [tex]\(-\frac{\left(\frac{\partial}{\partial x}\right)}{\left(\frac{\partial}{\partial y}\right)}\)[/tex].
- Substituting the values, we have:
[tex]\[
\text{slope} = -\frac{1}{-12} = 0.0833.
\][/tex]
6. Write the equation of the tangent line:
- The point-slope form of the equation of a line is:
[tex]\[
y - y_1 = m(x - x_1),
\][/tex]
where [tex]\((x_1, y_1)\)[/tex] is the point of tangency and [tex]\(m\)[/tex] is the slope.
- Plugging in [tex]\(m = 0.0833\)[/tex], [tex]\(x_1 = 1\)[/tex], and [tex]\(y_1 = 2\)[/tex], we get:
[tex]\[
y - 2 = 0.0833(x - 1).
\][/tex]
7. Simplify the equation:
- Expanding and rearranging gives:
[tex]\[
y - 2 = 0.0833x - 0.0833.
\][/tex]
- Adding 2 to both sides yields the final equation of the tangent line:
[tex]\[
y = 0.0833x + 1.9167.
\][/tex]
This is the equation of the tangent line to the curve at the point [tex]\((1, 2)\)[/tex].
1. Identify the given equation: The equation of the curve is
[tex]\[
y^4 - 2y^3 - 5y^2 - 6x^2 - x^3 + 23 = -4x^4.
\][/tex]
2. Rearrange the equation: For simplicity in differentiating, write this as:
[tex]\[
y^4 - 2y^3 - 5y^2 - 6x^2 - x^3 + 23 + 4x^4 = 0.
\][/tex]
3. Calculate partial derivatives:
- To find the slope of the tangent line, you need the derivatives with respect to [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
- The partial derivative with respect to [tex]\(x\)[/tex] is:
[tex]\[
\frac{\partial}{\partial x}\left(y^4 - 2y^3 - 5y^2 - 6x^2 - x^3 + 23 + 4x^4\right) = 16x^3 - 3x^2 - 12x.
\][/tex]
- The partial derivative with respect to [tex]\(y\)[/tex] is:
[tex]\[
\frac{\partial}{\partial y}\left(y^4 - 2y^3 - 5y^2 - 6x^2 - x^3 + 23 + 4x^4\right) = 4y^3 - 6y^2 - 10y.
\][/tex]
4. Evaluate the partial derivatives at the point [tex]\((1, 2)\)[/tex]:
- Substituting [tex]\(x = 1\)[/tex] and [tex]\(y = 2\)[/tex] into the partial derivatives, we get:
- [tex]\(\left. \frac{\partial}{\partial x} \right|_{(1, 2)} = 16(1)^3 - 3(1)^2 - 12(1) = 1\)[/tex].
- [tex]\(\left. \frac{\partial}{\partial y} \right|_{(1, 2)} = 4(2)^3 - 6(2)^2 - 10(2) = -12\)[/tex].
5. Calculate the slope of the tangent line:
- The slope of the tangent line for an implicit curve is given by [tex]\(-\frac{\left(\frac{\partial}{\partial x}\right)}{\left(\frac{\partial}{\partial y}\right)}\)[/tex].
- Substituting the values, we have:
[tex]\[
\text{slope} = -\frac{1}{-12} = 0.0833.
\][/tex]
6. Write the equation of the tangent line:
- The point-slope form of the equation of a line is:
[tex]\[
y - y_1 = m(x - x_1),
\][/tex]
where [tex]\((x_1, y_1)\)[/tex] is the point of tangency and [tex]\(m\)[/tex] is the slope.
- Plugging in [tex]\(m = 0.0833\)[/tex], [tex]\(x_1 = 1\)[/tex], and [tex]\(y_1 = 2\)[/tex], we get:
[tex]\[
y - 2 = 0.0833(x - 1).
\][/tex]
7. Simplify the equation:
- Expanding and rearranging gives:
[tex]\[
y - 2 = 0.0833x - 0.0833.
\][/tex]
- Adding 2 to both sides yields the final equation of the tangent line:
[tex]\[
y = 0.0833x + 1.9167.
\][/tex]
This is the equation of the tangent line to the curve at the point [tex]\((1, 2)\)[/tex].
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