# Determine A so that the curve y = 7x + 33 can be written in parametric form as x(t) = t - 5, y(t)...

## Question:

Determine {eq}A {/eq} so that the curve {eq}y = 7x + 33 {/eq} can be written in parametric form as

{eq}x(t) = t - 5, \quad y(t) = At - 2 {/eq}.

1. {eq}A = -5 {/eq}

2. {eq}A = 6 {/eq}

3. {eq}A = 7 {/eq}

4. {eq}A = 5 {/eq}

5. {eq}A = -7 {/eq}

6. {eq}A = -6 {/eq}

## Arbitrary constant for Parametric Equation:

Firstly, we'll compute the first differentiation of all the given equations with their respective variables respectively and then substitute all the obtained values in the general mathematical relation shown below for the required value of constant.

{eq}\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}} {/eq}

We are given:

The equation for a curve is:

{eq}y = 7x + 33 {/eq}

The parametric equations are:

{eq}x(t) = t - 5 \\ y(t) = At - 2 {/eq}

Firstly, we'll compute 1st derivative of the given curve with respect to {eq}x {/eq}.

{eq}\begin{align*} \displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}&=\frac{\mathrm{d} (7x + 33 )}{\mathrm{d} x}\\ &=(7(1) + 0 )\\ &=7 \\ \end{align*} {/eq}

Now, we'll compute the derivatives of the given parametric equations with respect to {eq}t {/eq} individually.

{eq}\begin{align*} \displaystyle \frac{\mathrm{d} x(t)}{\mathrm{d} t}&=\frac{\mathrm{d} ( t - 5 )}{\mathrm{d} t}\\ &=( 1- 0 )\\ &=1\\ \end{align*} {/eq}

{eq}\begin{align*} \displaystyle \frac{\mathrm{d} y(t)}{\mathrm{d} t}&=\frac{\mathrm{d} ( At - 2 )}{\mathrm{d} t}\\ &=( A(1)- 0)\\ &=A\\ \end{align*} {/eq}

By the derivative formula {eq}\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}} {/eq} with the above values, we have:

{eq}\begin{align*} \displaystyle 7&=\frac{A}{1}\\ \displaystyle A&=7\times 1\\ &=\boxed{7}\\ \end{align*} {/eq}

Thus, option (3) is correct.