Question

If t varies jointly with x and the square of v, and t=39 when x=square root of 10 and v=3.87, find t when x=square root of 3 and v=7.21

Answer 1

The variable t varies jointly with x and the square of v. We can use a constant of proportionality k and write:

$t=k\cdot x\cdot\sqrt[]{v}$

We know that t = 39 when x = √10 and v = 3.87.

We can use this to find the value of k first:

$\begin{gathered} t=k\cdot x\cdot\sqrt[]{v} \\ k=\frac{t}{x\sqrt[]{v}} \\ k=\frac{39}{\sqrt[]{10}\cdot\sqrt[]{3.87}} \\ k\approx\frac{39}{3.1623\cdot1.9672} \\ k\approx\frac{39}{6.2209} \\ k\approx6.26916 \end{gathered}$

Then, we can now calculate the value of t when x = √3 and v = 7.21 as:

$\begin{gathered} t=6.26916\cdot x\cdot\sqrt[]{v} \\ t=6.26916\cdot\sqrt[]{3}\cdot\sqrt[]{7.21} \\ t\approx6.26916\cdot1.7321\cdot2.6851 \\ t\approx29.1566 \end{gathered}$

Answer: t = 29.1566 (rounded to 4 decimals)