Math  /  Calculus

Questionhomework4.8: Problem 6 (1 point)
Find parametric equations for the tangent line at t=3t=3 for the motion of a particle given by x(t)=8t2+7,y(t)=1t3x(t)=8 t^{2}+7, y(t)=1 t^{3}. For the line, x(t)=y(t)=\begin{array}{l} x(t)=\square \\ y(t)=\square \end{array} (Note that because the correctness of a parametrically described line depends on both x(t)x(t) and y(t)y(t), both of your answers may be marked incorrect if there is an error in one of them.)
Note: You can earn partial credit on this problem.

Studdy Solution
Write the parametric equations for the tangent line using the point (79,27) (79, 27) and the velocity vector (48,27) (48, 27) .
The parametric equations for the tangent line are: x(t)=79+48(t3) x(t) = 79 + 48(t - 3) y(t)=27+27(t3) y(t) = 27 + 27(t - 3)
Simplify the equations: x(t)=79+48t144=48t65 x(t) = 79 + 48t - 144 = 48t - 65 y(t)=27+27t81=27t54 y(t) = 27 + 27t - 81 = 27t - 54
The parametric equations for the tangent line are:
x(t)=48t65 \boxed{x(t) = 48t - 65} y(t)=27t54 \boxed{y(t) = 27t - 54}

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