Find the equation of the tangent line to $\newline$ $x=\sqrt[3]{t}$ and $y=\frac{1}{2}(t^{2}-2)$ at the point $(2,31)$

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Algebra 2

Quadratic equation with complex roots

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Q. Find the equation of the tangent line to

\newline

x=\sqrt[3]{t}

and

y=\frac{1}{2}(t^{2}-2)

at the point

(2,31)

Find Derivative: To find the equation of the tangent line, we need to find the derivative of $y$ with respect to $x$ . This can be done by finding $\frac{dy}{dt}$ and $\frac{dx}{dt}$ and then dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$ to get $\frac{dy}{dx}$ .
Calculate $\frac{dx}{dt}$ : First, let's find $\frac{dx}{dt}$ . Given $x = \sqrt[3]{t}$ , we differentiate with respect to $t$ to get $\frac{dx}{dt} = \frac{1}{3}t^{-\frac{2}{3}}$ .
Calculate $\frac{dy}{dt}$ : Now, let's find $\frac{dy}{dt}$ . Given $y = \frac{1}{2}(t^2 - 2)$ , we differentiate with respect to $t$ to get $\frac{dy}{dt} = t$ .
Find $\frac{dy}{dx}$ : Next, we find $\frac{dy}{dx}$ by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$ . So, $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{t}{\left(\frac{1}{3}t^{-\frac{2}{3}}\right)} = 3t^{\frac{5}{3}}$ .
Find t Value: We need to find the value of $t$ that corresponds to the point $(2,31)$ . Since $x = \sqrt[3]{t}$ , we solve the equation $2 = \sqrt[3]{t}$ to find $t$ . Cubing both sides gives us $t^3 = 8$ , so $t = 2$ .
Calculate Slope: Now we substitute $t = 2$ into the derivative $\frac{dy}{dx}$ to find the slope of the tangent line at the point $(2,31)$ . The slope $m$ is $3*(2)^{\frac{5}{3}}$ .
Use Point-Slope Form: Calculating the slope $m$ , we get $m = 3\cdot(2)^{\frac{5}{3}} = 3\cdot(2^{\frac{5}{3}}) = 3\cdot(2^{2}\cdot2^{\frac{1}{3}}) = 3\cdot4\cdot\sqrt[3]{2} = 12\cdot\sqrt[3]{2}$ .
Substitute Values: With the slope $m = 12\sqrt[3]{2}$ and the point $(2,31)$ , we can use the point-slope form of the equation of a line to find the equation of the tangent line: $y - y_1 = m(x - x_1)$ , where $(x_1, y_1)$ is the point $(2,31)$ .
Simplify Equation: Substituting the values into the point-slope form, we get $y - 31 = 12\sqrt[3]{2}(x - 2)$ .
Final Tangent Line Equation: To simplify, we distribute the slope on the right side of the equation: $y - 31 = 12\sqrt[3]{2}x - 24\sqrt[3]{2}$ .
Final Tangent Line Equation: To simplify, we distribute the slope on the right side of the equation: $y - 31 = 12\sqrt[3]{2}x - 24\sqrt[3]{2}$ . Finally, we add $31$ to both sides to get the equation of the tangent line in slope-intercept form: $y = 12\sqrt[3]{2}x - 24\sqrt[3]{2} + 31$ .