y(1)=1quadx^(')(t)=3quady^(')(t)=2t Write the equation of the line tangent to the curve given by x(t)=3t-1 and y(t)=t^(2) at the point where t=1.

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y(1)=1quadx^(')(t)=3quady^(')(t)=2t Write the equation of the line tangent to the curve given by  x(t)=3t-1 and  y(t)=t^(2) at the point where  t=1.

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Find Slope of Tangent: To find the equation of the tangent line, we need to determine the slope of the tangent line at the point where t=1. The slope of the tangent line is given by the derivative of y with respect to x, which can be found using the derivatives of y with respect to t and x with respect to t.Derivative of x: First, we find the derivative of x with respect to t, denoted as x'(t). Given x(t)=3t-1, we differentiate with respect to t to get x'(t)=3.Derivative of y: Next, we find the derivative of y with respect to t, denoted as y'(t). Given y(t)=t^2, we differentiate with respect to t to get y'(t)=2t.Evaluate Derivatives at t=1: Now, we evaluate both derivatives at t=1. For x'(1), we have x'(1)=3. For y'(1), we have y'(1)=2*1=2.Calculate Tangent Line Slope: The slope of the tangent line at t=1 is the ratio of the derivatives, which is (y'(1))/(x'(1)) = 2/3.Find Coordinates at t=1: We also need the coordinates of the point on the curve at t=1. We substitute t=1 into the original equations to find the coordinates. For x(1), we have x(1)=3*1-1=2. For y(1), we have y(1)=(1)^2=1.Use Point-Slope Form: With the slope of the tangent line and the point of tangency, we can use the point-slope form of the equation of a line: y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point of tangency.Substitute Slope and Point: Substituting the slope and the point into the point-slope form, we get y - 1 = (2/3)(x - 2).Convert to Slope-Intercept Form: To write the equation in slope-intercept form, we can distribute the slope and simplify: y - 1 = (2/3)x - (4/3).Solve for y: Finally, we add 1 to both sides to solve for y: y = (2/3)x - (4/3) + 1.Final Tangent Line Equation: Simplifying the equation, we combine like terms: y = (2/3)x - (4/3) + (3/3).The final equation of the tangent line is y = (2/3)x - (1/3).

$y(1)=1\quad x'(t)=3\quad y'(t)=2t$ $\newline$ Write the equation of the line tangent to the curve given by $\newline$ $x(t)=3t-1$ and $\newline$ $y(t)=t^{2}$ at the point where $\newline$ $t=1$ .

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y(1)=1x′(t)=3y′(t)=2ty(1)=1\quad x'(t)=3\quad y'(t)=2ty(1)=1x′(t)=3y′(t)=2t\newlineWrite the equation of the line tangent to the curve given by \newlinex(t)=3t−1x(t)=3t-1x(t)=3t−1 and \newliney(t)=t2y(t)=t^{2}y(t)=t2 at the point where \newlinet=1t=1t=1.

$y(1)=1\quad x'(t)=3\quad y'(t)=2t$ $\newline$ Write the equation of the line tangent to the curve given by $\newline$ $x(t)=3t-1$ and $\newline$ $y(t)=t^{2}$ at the point where $\newline$ $t=1$ .