Parametric equations and a value for the parameter t are given x = (80cos 45^o)t, y = 6

Parametric equations and a value for the parameter t are given $$\displaystyle{x}={\left({80}{\cos{{45}}}^{{o}}\right)}{t},{y}={6}+{\left({80}{\sin{{45}}}^{{o}}\right)}{t}-{16}{t}^{{2}}$$. t = 2.
Find the coordinates of the point on the plane curve described by the parametric equations corresponding to the given value of t.

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Theodore Schwartz

Solution:

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

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

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

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

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

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