0
$\begingroup$

The question I've been given is $$\begin{array}{c|c}t&v\\\hline3&38\\12&200\end{array}$$ Modelling equation is $$v=k\sqrt{t-a}$$ Calculate $a$ and $k$.

I tried to solve like:

$$38 = k \sqrt{12-a}$$ $$200 = k \sqrt{3-a}$$

$$162 = k \sqrt{12-a} - k \sqrt{3-a}$$

$$\frac{162}k = \sqrt{12-a} - \sqrt{3-a}$$

$$(162/k)^2 = 12-a - 3 + a$$

$$(162/k)^2 = 9$$

$$162/k = 3$$

$$162/3 = k$$

$$k = 54$$

WHICH IS GIVEN AS WRONG. $k$ should be $65.45$

$\endgroup$
  • $\begingroup$ mathjax resources to learn how to typeset maths. $\endgroup$ – Siong Thye Goh Jul 22 '18 at 4:54
  • $\begingroup$ You swapped $3$ and $12$ ! $\endgroup$ – Yves Daoust Jul 23 '18 at 11:50
0
$\begingroup$

It seems to me that in general this question would be easier if you were to square both sides:

$$ v^2=k^2(t-a) $$

With the two $(t, v)$ pairs of $(3, 38)$ and $(12, 200)$, we have

$$ 1444 = 3k^2-ak^2 \\ 40000 = 12k^2-ak^2 $$

Subtracting the former from the latter gives us

$$ 9k^2 = 38556 \\ k^2 = 4284 k = \sqrt{4284} \approx 65.452 $$

Then substituting $k^2 = 4284$ back into either of the above yields

$$ a = 12-\frac{40000}{4284} = 3-\frac{1444}{4284} \approx 2.663 $$

Assuming that this model is supposed to produce real numbers as output, the implication is that it cannot be used for $t < a$.

$\endgroup$
  • $\begingroup$ Thanks, that got it $\endgroup$ – DJx Jul 22 '18 at 5:59
0
$\begingroup$

First mistake:

  • When time is $3$, the volume is $38$.

  • When time is $12$, the volume is $200$.

  • You have made the mistake of thinking when time is $3$, the volume is $200$. Similary for the other equation.

Second mistake:

  • In general, $(\sqrt{a} - \sqrt{b})^2 \ne a - b$.

Guide:

  • You might like to divide the equations rather than subtract the equations.
$\endgroup$
  • $\begingroup$ Thanks for the reminder that (a−−√−b√)2≠a−b. I forgot that $\endgroup$ – DJx Jul 22 '18 at 5:59
0
$\begingroup$

Using your data, you have in fact $$38 = k \sqrt{3-a}\tag 1$$ $$200 = k \sqrt{12-a}\tag 2$$ Make the ratio $$\frac{38}{200}=\frac{\sqrt{3-a} } {\sqrt{12-a} }$$ Square both sides to get $$\frac{361}{10000}=\frac{3-a}{12-a}\implies a=\frac{2852}{1071}$$ Plug in $(1)$ to get $k=6 \sqrt{119}\approx 65.4523$.

$\endgroup$
0
$\begingroup$

Rewrite the model as

$$k^2t-k^2a=v^2$$

and you have a system of two linear equations in the unknowns $k^2$ and $k^2a$.

$$\begin{cases}3k^2-k^2a&=38^2,\\12k^2-k^2a&=200^2.\end{cases}$$

Then solving the system

$$k^2= 4284,$$ $$k^2a=11408.$$

The rest is easy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.