If $$ r $$ varies direct as the square of $$ t $$ and $$ r=216 $$ when $$ t=6 $$, find $$ + \text { when } r=512 $$

Question

UpStudy AI · Accepted Answer

Since $$ r $$ varies directly as the square of $$ t $$, we can express this relationship mathematically as:

$$
r = k t^2
$$

where $$ k $$ is a constant.

Given that $$ r = 216 $$ when $$ t = 6 $$, we can substitute these values into the equation to find $$ k $$:

$$
216 = k (6^2)
$$

Calculating $$ 6^2 $$:

$$
6^2 = 36
$$

Now substituting this back into the equation:

$$
216 = k \cdot 36
$$

To solve for $$ k $$, we divide both sides by 36:

$$
k = \frac{216}{36} = 6
$$

Now that we have $$ k $$, we can express the relationship between $$ r $$ and $$ t $$ as:

$$
r = 6 t^2
$$

Next, we need to find $$ t $$ when $$ r = 512 $$:

$$
512 = 6 t^2
$$

To isolate $$ t^2 $$, we divide both sides by 6:

$$
t^2 = \frac{512}{6} = \frac{256}{3}
$$

Now, we take the square root of both sides to find $$ t $$:

$$
t = \sqrt{\frac{256}{3}} = \frac{\sqrt{256}}{\sqrt{3}} = \frac{16}{\sqrt{3}}
$$

To rationalize the denominator, we multiply the numerator and denominator by $$ \sqrt{3} $$:

$$
t = \frac{16 \sqrt{3}}{3}
$$

Thus, the value of $$ t $$ when $$ r = 512 $$ is:

$$
\boxed{\frac{16 \sqrt{3}}{3}}
$$
The value of $$ t $$ when $$ r = 512 $$ is $$ \frac{16 \sqrt{3}}{3} $$.

If \( r \) varies direct as the square of \( t \) and \( r=216 \) when \( t=6 \), find \[ + \text { when } r=512 \]