Practice Problems
\(\textbf{1)}\) Express y in terms of x.
\(x=3t\)
\(y=t+2\)
The answer is \( y=\frac{1}{3}x+2 \)
\(\,\,\,\,\,\,\text{Express t in terms of }x:\, x=3t \,\,\, \Rightarrow \,\,\, t= \frac{x}{3}\)
\(\,\,\,\,\,\,\text{Substitute } \frac{x}{3} \text{ in for } t \text{ in the other equation}:\, y=t+2 \,\,\,\Rightarrow \,\,\, y=\left(\frac{x}{3}\right)+2\)
\(\,\,\,\,\,\,\text{The answer is } y=\frac{1}{3}x+2 \)
\(\textbf{2)}\) Express y in terms of x.
\(x=5t\)
\(y=t^2\)
The answer is \( y=\displaystyle\frac{x^2}{25} \)
\(\,\,\,\,\,\,\text{Express t in terms of }x:\, x=5t \,\,\, \Rightarrow \,\,\, t= \frac{x}{5}\)
\(\,\,\,\,\,\,\text{Substitute } \frac{x}{5} \text{ in for } t \text{ in the other equation}:\, y=t^2 \,\,\, \Rightarrow \,\,\, y=\left(\frac{x}{5}\right)^2\)
\(\,\,\,\,\,\,\text{The answer is } y=\displaystyle\frac{x^2}{25} \)
\(\textbf{3)}\) Express y in terms of x.
\(x=t-2\)
\(y=t^2-1\)
The answer is \( y=x^2+4x+3 \)
\(\,\,\,\,\,\,\text{Express t in terms of }x:\, x=t-2 \,\,\, \Rightarrow \,\,\, t= x+2\)
\(\,\,\,\,\,\,\text{Substitute } x+2 \text{ in for } t \text{ in the other equation}:\, y=t^2-1 \,\,\, \Rightarrow \,\,\, y=\left(x+2\right)^2-1\)
\(\,\,\,\,\,\,y=\left(x^2+4x+4\right)-1\)
\(\,\,\,\,\,\,\text{The answer is } y=x^2+4x+3 \)
Challenge Questions
\(\textbf{4)}\) Express in terms of x and y.
\(x=\cos{t}\)
\(y=\sin{t}\)
The answer is \( x^2+y^2=1 \)
\(\,\,\,\,\,\,x=\cos t \,\,\, \Rightarrow \,\,\, x^2=\cos^2 t\)
\(\,\,\,\,\,\,y=\sin t \,\,\, \Rightarrow \,\,\, y^2=\sin^2 t\)
\(\,\,\,\,\,\,\text{Recall the pythagorean property} \sin^2 t + \cos^2 t =1\)
\(\,\,\,\,\,\,\text{Substitute from above } x^2 + y^2 =1\)
\(\,\,\,\,\,\,\text{The answer is } x^2+y^2=1 \)
\(\textbf{5)}\) Express in terms of x and y.
\(x=4\cos{t}\)
\(y=3\sin{t}\)
The answer is \( \left(\frac{x}{4}\right)^2 +\left(\frac{y}{3}\right)^2=1 \)
\(\,\,\,\,\,\,x=4\cos t \,\,\, \Rightarrow \,\,\, \frac{x}{4}=\cos t \,\,\, \Rightarrow \,\,\, \left(\frac{x}{4}\right)^2=\cos^2 t\)
\(\,\,\,\,\,\,y=3\sin t \,\,\, \Rightarrow \,\,\, \frac{y}{3}=\sin t \,\,\, \Rightarrow \,\,\, \left(\frac{y}{3}\right)^2=\sin^2 t\)
\(\,\,\,\,\,\,\text{Recall the pythagorean property} \sin^2 t + \cos^2 t =1\)
\(\,\,\,\,\,\,\text{Substitute from above } \left(\frac{x}{4}\right)^2 +\left(\frac{y}{3}\right)^2=1\)
\(\,\,\,\,\,\,\text{The answer is } \left(\frac{x}{4}\right)^2 +\left(\frac{y}{3}\right)^2=1 \)
See Related Pages\(\)
