Ex. 4 Find the equation of the tangent line to x=root(3)(t) and y=(1)/(2)(t^(2)-2) at the point (2,31)

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Ex. 4 Find the equation of the tangent line to  x=root(3)(t) and  y=(1)/(2)(t^(2)-2) at the point  (2,31)

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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 dy/dt and dx/dt and then dividing dy/dt by dx/dt to get dy/dx.Calculate dx/dt: First, let's find dx/dt. Given x = root(3)(t), we differentiate with respect to t to get dx/dt = (1/3)t^(-2/3).Calculate dy/dt: Now, let's find dy/dt. Given y = (1/2)(t^2 - 2), we differentiate with respect to t to get dy/dt = t.Find dy/dx: Next, we find dy/dx by dividing dy/dt by dx/dt. So, dy/dx = (dy/dt) / (dx/dt) = (t) / ((1/3)t^(-2/3)) = 3t^(5/3).Find t Value: We need to find the value of t that corresponds to the point (2,31). Since x = root(3)(t), we solve the equation 2 = root(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 dy/dx to find the slope of the tangent line at the point (2,31). The slope m is 3*(2)^(5/3).Use Point-Slope Form: Calculating the slope m, we get m = 3*(2)^(5/3) = 3*(2^(5/3)) = 3*(2^(2)*2^(1/3)) = 3*4*root(3)(2) = 12*root(3)(2).Substitute Values: With the slope m = 12*root(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 - y1 = m(x - x1), where (x1, y1) is the point (2,31).Simplify Equation: Substituting the values into the point-slope form, we get y - 31 = 12*root(3)(2)(x - 2).Final Tangent Line Equation: To simplify, we distribute the slope on the right side of the equation: y - 31 = 12*root(3)(2)x - 24*root(3)(2).Finally, we add 31 to both sides to get the equation of the tangent line in slope-intercept form: y = 12*root(3)(2)x - 24*root(3)(2) + 31.

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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Find the equation of the tangent line to \newlinex=t3x=\sqrt[3]{t}x=3t​ and y=12(t2−2)y=\frac{1}{2}(t^{2}-2)y=21​(t2−2) at the point (2,31)(2,31)(2,31)

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