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Sagot :
To find the equation of the tangent line to the curve [tex]\( x^{2/3} + y^{2/3} = 4 \)[/tex] at the point [tex]\((-3\sqrt{3}, 1)\)[/tex], we will use implicit differentiation and find the slope at the given point, and then form the equation of the tangent line.
### Step-by-Step Solution:
1. Implicit Differentiation:
Begin by differentiating the given equation [tex]\( x^{2/3} + y^{2/3} = 4 \)[/tex] implicitly with respect to [tex]\( x \)[/tex].
[tex]\[ \frac{d}{dx}(x^{2/3} + y^{2/3}) = \frac{d}{dx}(4) \][/tex]
Differentiating term by term:
[tex]\[ \frac{d}{dx}(x^{2/3}) + \frac{d}{dx}(y^{2/3}) = 0 \][/tex]
Using the chain rule for the second term:
[tex]\[ \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0 \][/tex]
2. Solve for [tex]\(\frac{dy}{dx}\)[/tex] (Slope of Tangent Line):
Isolate [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \frac{2}{3}x^{-1/3} = -\frac{2}{3}y^{-1/3} \frac{dy}{dx} \][/tex]
Simplify:
[tex]\[ x^{-1/3} = -y^{-1/3} \frac{dy}{dx} \][/tex]
Divide both sides by [tex]\( -y^{-1/3} \)[/tex]:
[tex]\[ \frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} \][/tex]
Rewrite it with positive exponents:
[tex]\[ \frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}} \][/tex]
3. Calculate the Slope at the Given Point [tex]\( (-3\sqrt{3}, 1) \)[/tex]:
Substitute [tex]\( x = -3\sqrt{3} \)[/tex] and [tex]\( y = 1 \)[/tex] into the slope expression:
[tex]\[ \left.\frac{dy}{dx}\right|_{(-3\sqrt{3}, 1)} = -\frac{(1)^{1/3}}{(-3\sqrt{3})^{1/3}} \][/tex]
Simplify the slope calculation:
[tex]\[ (1)^{1/3} = 1 \][/tex]
[tex]\[ (-3\sqrt{3})^{1/3} = -\sqrt[3]{27} = -3^{1/3} \cdot (\sqrt{3})^{1/3} = -3 \cdot \sqrt[3]{3} \][/tex]
Therefore, the slope is:
[tex]\[ \left.\frac{dy}{dx}\right|_{(-3\sqrt{3}, 1)} = -\frac{1}{-3\sqrt[3]{3}} = \frac{1}{3\sqrt[3]{3}} \][/tex]
4. Equation of the Tangent Line:
The equation of a tangent line in point-slope form is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Where [tex]\( (x_1, y_1) \)[/tex] is the point [tex]\( (-3\sqrt{3}, 1) \)[/tex] and [tex]\( m \)[/tex] is the slope calculated above:
[tex]\[ y - 1 = \frac{1}{3\sqrt[3]{3}}(x + 3\sqrt{3}) \][/tex]
Simplify the equation:
[tex]\[ y - 1 = \frac{1}{3\sqrt[3]{3}} x + \frac{1}{3\sqrt[3]{3}} \cdot 3\sqrt{3} \][/tex]
Since [tex]\( \frac{1}{3\sqrt[3]{3}} \cdot 3\sqrt{3} = 1 \)[/tex]:
[tex]\[ y - 1 = \frac{1}{3\sqrt[3]{3}} x + 1 \][/tex]
Add 1 to both sides:
[tex]\[ y = \frac{1}{3\sqrt[3]{3}} x + 2 \][/tex]
Thus, the equation of the tangent line to the curve [tex]\( x^{2/3} + y^{2/3} = 4 \)[/tex] at the point [tex]\((-3\sqrt{3}, 1)\)[/tex] is:
[tex]\[ y = \frac{1}{3\sqrt[3]{3}} x + 2 \][/tex]
### Step-by-Step Solution:
1. Implicit Differentiation:
Begin by differentiating the given equation [tex]\( x^{2/3} + y^{2/3} = 4 \)[/tex] implicitly with respect to [tex]\( x \)[/tex].
[tex]\[ \frac{d}{dx}(x^{2/3} + y^{2/3}) = \frac{d}{dx}(4) \][/tex]
Differentiating term by term:
[tex]\[ \frac{d}{dx}(x^{2/3}) + \frac{d}{dx}(y^{2/3}) = 0 \][/tex]
Using the chain rule for the second term:
[tex]\[ \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0 \][/tex]
2. Solve for [tex]\(\frac{dy}{dx}\)[/tex] (Slope of Tangent Line):
Isolate [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \frac{2}{3}x^{-1/3} = -\frac{2}{3}y^{-1/3} \frac{dy}{dx} \][/tex]
Simplify:
[tex]\[ x^{-1/3} = -y^{-1/3} \frac{dy}{dx} \][/tex]
Divide both sides by [tex]\( -y^{-1/3} \)[/tex]:
[tex]\[ \frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} \][/tex]
Rewrite it with positive exponents:
[tex]\[ \frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}} \][/tex]
3. Calculate the Slope at the Given Point [tex]\( (-3\sqrt{3}, 1) \)[/tex]:
Substitute [tex]\( x = -3\sqrt{3} \)[/tex] and [tex]\( y = 1 \)[/tex] into the slope expression:
[tex]\[ \left.\frac{dy}{dx}\right|_{(-3\sqrt{3}, 1)} = -\frac{(1)^{1/3}}{(-3\sqrt{3})^{1/3}} \][/tex]
Simplify the slope calculation:
[tex]\[ (1)^{1/3} = 1 \][/tex]
[tex]\[ (-3\sqrt{3})^{1/3} = -\sqrt[3]{27} = -3^{1/3} \cdot (\sqrt{3})^{1/3} = -3 \cdot \sqrt[3]{3} \][/tex]
Therefore, the slope is:
[tex]\[ \left.\frac{dy}{dx}\right|_{(-3\sqrt{3}, 1)} = -\frac{1}{-3\sqrt[3]{3}} = \frac{1}{3\sqrt[3]{3}} \][/tex]
4. Equation of the Tangent Line:
The equation of a tangent line in point-slope form is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Where [tex]\( (x_1, y_1) \)[/tex] is the point [tex]\( (-3\sqrt{3}, 1) \)[/tex] and [tex]\( m \)[/tex] is the slope calculated above:
[tex]\[ y - 1 = \frac{1}{3\sqrt[3]{3}}(x + 3\sqrt{3}) \][/tex]
Simplify the equation:
[tex]\[ y - 1 = \frac{1}{3\sqrt[3]{3}} x + \frac{1}{3\sqrt[3]{3}} \cdot 3\sqrt{3} \][/tex]
Since [tex]\( \frac{1}{3\sqrt[3]{3}} \cdot 3\sqrt{3} = 1 \)[/tex]:
[tex]\[ y - 1 = \frac{1}{3\sqrt[3]{3}} x + 1 \][/tex]
Add 1 to both sides:
[tex]\[ y = \frac{1}{3\sqrt[3]{3}} x + 2 \][/tex]
Thus, the equation of the tangent line to the curve [tex]\( x^{2/3} + y^{2/3} = 4 \)[/tex] at the point [tex]\((-3\sqrt{3}, 1)\)[/tex] is:
[tex]\[ y = \frac{1}{3\sqrt[3]{3}} x + 2 \][/tex]
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