Direct Variation

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Question 1 of 3

1. Question
Which of the following is a direct variation?
- 1. $y=3x$
- 2. $y=3/x$
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Direct Variation

A relationship between two variables, where if one increases the other also increases. Similarly, if one variable decreases the other also decreases.

First, list down a few values for $x$ and substitute it to the equations to compute for the corresponding $y$ values

First equation:

$x$ $1$ $2$ $3$

$y$

Second equation:

$x$ $1$ $2$ $3$

$y$

$y$ $=$ $3$ $x$

$y$ $=$ $\frac{3}{\color{#004ec4}{x}}$

$y$ $=$ $3$ $(1)$

$y$ $=$ $3$

$y$ $=$ $\frac{3}{\color{#004ec4}{1}}$

$y$ $=$ $3$

$y$ $=$ $3$ $(2)$

$y$ $=$ $6$

$y$ $=$ $\frac{3}{\color{#004ec4}{2}}$

$y$ $=$ $3/2$

$y$ $=$ $3$ $(3)$

$y$ $=$ $9$

$y$ $=$ $\frac{3}{\color{#004ec4}{3}}$

$y$ $=$ $1$

$x$ $1$ $2$ $3$

$y$ $3$ $6$ $9$

$x$ $1$ $2$ $3$

$y$ $3$ $3/2$ $1$

Notice that on the first equation, as the $x$ value increases, the $y$ value also increases.

Hence, the first equation, ( $y=3x$ ) is a direct variation.

It is also worth noting that in the second equation, as the $x$ value increases, the $y$ value decreases. This is an inverse variation.

$y=3x$
Question 2 of 3

2. Question
A car travels $600$ km in $8$ hours. How far would it travel in $11$ hours, given that this is a direct variation?
- (825) $km$
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Direct Variation Formula

$y$ $=k$ $x$

where $k$ is the constant of variation

Remember

A direct variation is a relationship between two variables, where if one increases the other also increases. Similarly, if one variable decreases the other also decreases.

First, solve for $k$ , the constant of variation, by plugging in the known values to the Direct Variation Formula.

$y$ $=$ $600$ km

$x$ $=$ $8$ hours

$y$ $=$ $k$ $x$ Direct Variation Formula

$600$ $=$ $k$ $(8)$ Substitute known values

$600$ $=$ $8k$

$600$ $divide 8$ $=$ $8k$ $divide 8$ Divide both sides by $8$

$75$ $=$ $k$

$k$ $=$ $75$

Next, rewrite the Direct Variation Formula with $k$ substituted.

$y$ $=$ $kx$

$y$ $=$ $75x$ Substitute $k$

Finally, use the new formula and substitute $11$ hours

$y$ $=$ $75x$ New formula

$y$ $=$ $75(11)$ Substitute $11$ hours

$y$ $=$ $825$

Hence, the car would travel $825$ km in $11$ hours

$825$ $km$
Question 3 of 3

3. Question
The time taken, $T$ , for a pendulum to swing varies directly at the square root of its length. If one swing of a $100$ -centimetre pendulum takes $2$ seconds, find the time taken for one swing of a $36$ centimetre pendulum.

Write answer in decimal form
- (1.2) seconds
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Direct Variation Formula

$y$ $=k$ $x$

where $k$ is the constant of variation

Remember

A direct variation is a relationship between two variables, where if one increases the other also increases. Similarly, if one variable decreases the other also decreases.

First, rewrite the Direct Variation Formula according to the problem given.

$T$ varies directly at the square root of its length

$T$ $=$ $ksqrtL$ Insert $k$ , the constant of variation

Now, solve for $k$ by plugging in the known values to the new formula.

$T$ $=$ $2$ seconds

$L$ $=$ $100$ cm

$T$ $=$ $k$ $sqrtL$ New formula

$2$ $=$ $k$ $sqrt100$ Substitute known values

$2$ $=$ $10k$

$2$ $divide 10$ $=$ $10k$ $divide 10$ Divide both sides by $10$

$0.2$ $=$ $k$

$k$ $=$ $0.2$

Add in the value of $k$ to the new formula.

$T$ $=$ $ksqrtL$

$T$ $=$ $0.2sqrtL$ Substitute $k$

Finally, use the updated formula and substitute $36$ cm to the length

$T$ $=$ $0.2sqrtL$ Updated formula

$T$ $=$ $0.2sqrt36$ Substitute $36$ cm

$T$ $=$ $0.2(6)$

$T$ $=$ $1.2$

Hence, the $36$ centimetre pendulum would take $1.2$ seconds to do one full swing.

$1.2$ seconds

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Quiz summary

Information

1. Question

Hint

Nice Job!

Direct Variation

2. Question

Hint

Excellent!

Direct Variation Formula

Remember

3. Question

Hint

Fantastic!

Direct Variation Formula

Remember

Quizzes

$y$	$=$	$k$ $x$	Direct Variation Formula
$600$	$=$	$k$ $(8)$	Substitute known values
$600$	$=$	$8k$
$600$ $divide 8$	$=$	$8k$ $divide 8$	Divide both sides by $8$
$75$	$=$	$k$
$k$	$=$	$75$

$y$	$=$	$75x$	New formula
$y$	$=$	$75(11)$	Substitute $11$ hours
$y$	$=$	$825$

$T$	varies directly	at the square root of its length
$T$	$=$	$ksqrtL$	Insert $k$ , the constant of variation

$T$	$=$	$k$ $sqrtL$	New formula
$2$	$=$	$k$ $sqrt100$	Substitute known values
$2$	$=$	$10k$
$2$ $divide 10$	$=$	$10k$ $divide 10$	Divide both sides by $10$
$0.2$	$=$	$k$
$k$	$=$	$0.2$

$T$	$=$	$0.2sqrtL$	Updated formula
$T$	$=$	$0.2sqrt36$	Substitute $36$ cm
$T$	$=$	$0.2(6)$
$T$	$=$	$1.2$

$x$	$1$	$2$	$3$
$y$

$x$	$1$	$2$	$3$
$y$

$x$	$1$	$2$	$3$
$y$	$3$	$6$	$9$

$x$	$1$	$2$	$3$
$y$	$3$	$3/2$	$1$