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Maiclubk

Answered question

2021-08-22

Parametric equations and a value for the parameter t are given

x = (80 {\cos 45}^{o}) t, y = 6 + (80 {\sin 45}^{o}) t - 16 t^{2}

. t = 2.
Find the coordinates of the point on the plane curve described by the parametric equations corresponding to the given value of t.

Theodore Schwartz

Skilled2021-08-23Added 99 answers

Solution:

1.For t=2, $x = (80 \cos 45^{o}) t = (80 \cos 45^{o}) (2) = 80 \cdot \frac{\sqrt{2}}{2} \cdot 2 = 80 \sqrt{2}$

For t=2, $y = 6 + (80 \sin 45^{o}) t - 16$

$t^{2} = 6 + (80 \sin 45) (2) = 16 {(2)}^{2} = 6 (80 \cdot \frac{\sqrt{2}}{2} \cdot 2) - 16 (4) = 6 + 80 \sqrt{2} - 64 = 80 \sqrt{2} - 58$

2. So the coordinate of the point is=(x,y)=

$(80 \sqrt{2}, 8 - \sqrt{2} - 58)$

Answer: $(80 \sqrt{2}, 8 - \sqrt{2} - 58)$

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