Math / Algebra

QuestionIn 2004, it was reported that "the relationship between body mass, $M$ , and standard metabolic rate, $B$ , among living organisms remains controversial. However, in many cases $B$ is approximately proportional to the three-quarters power of $M$ ." (a) Write a function that represents this relationship. If you must introduce a new variable, use $k$ . $B=$ $\square$ K(M3(4)) (b) The average mass of an African forest elephant is 4.9 metric tons, and that of a typical mouse is 25 grams. Use part (a) to determine how many times greater the metabolic rate of an elephant is than that of a mouse. Recall that 1 metric ton = 1,000,000 grams. $\square$ times greater

Studdy Solution

STEP 1

What is this asking? We need to write a formula relating metabolic rate to body mass, and then use it to compare the metabolic rates of an elephant and a mouse. Watch out! The units for the elephant and mouse masses are different!
We need to make sure to convert them to the same units before comparing.
Also, "proportional to" doesn't mean "equal to"!

STEP 2

1. Write the function
2. Calculate the elephant's metabolic rate
3. Calculate the mouse's metabolic rate
4. Compare the metabolic rates

STEP 3

" $B$ is approximately proportional to the three-quarters power of $M$ " means that $B$ is almost equal to $M$ raised to the power of $\frac{3}{4}$ , multiplied by some constant.
Let's call that constant $k$ .

STEP 4

So, our function is $B = k \cdot M^{\frac{3}{4}}$ .
This captures the idea that as $M$ increases, $B$ also increases, but not at the same rate.
It's raised to the power of three-quarters!

STEP 5

The elephant's mass is 4.9 metric tons.
Since 1 metric ton = 1,000,000 grams, the elephant's mass in grams is $4.9 \cdot 1{,}000{,}000 = 4{,}900{,}000$ grams.

STEP 6

Plugging this mass into our function, we get $B_{\text{elephant}} = k \cdot (4{,}900{,}000)^{\frac{3}{4}}$ .
We don't know $k$ , but don't worry, it will disappear when we compare the rates.

STEP 7

The mouse's mass is 25 grams.

STEP 8

Plugging this into our function, we get $B_{\text{mouse}} = k \cdot (25)^{\frac{3}{4}}$ .

STEP 9

We want to know how many times greater the elephant's metabolic rate is than the mouse's.
This means we need to divide the elephant's rate by the mouse's rate: $\frac{B_{\text{elephant}}}{B_{\text{mouse}}} = \frac{k \cdot (4{,}900{,}000)^{\frac{3}{4}}}{k \cdot (25)^{\frac{3}{4}}}$

STEP 10

Notice that the $k$ values divide to one, so we're left with: $\frac{(4{,}900{,}000)^{\frac{3}{4}}}{(25)^{\frac{3}{4}}}$

STEP 11

We can rewrite this as: $\left( \frac{4{,}900{,}000}{25} \right)^{\frac{3}{4}} = (196{,}000)^{\frac{3}{4}}$

STEP 12

Now, we calculate the final result: $(196{,}000)^{\frac{3}{4}} \approx 1{,}825.74 \approx 1{,}826$

STEP 13

$B = k \cdot M^{\frac{3}{4}}$ .
The elephant's metabolic rate is approximately 1,826 times greater than the mouse's.

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Write the function
2. Calculate the elephant's metabolic rate
3. Calculate the mouse's metabolic rate
4. Compare the metabolic rates

STEP 3

" $B$ is approximately proportional to the three-quarters power of $M$ " means that $B$ is almost equal to $M$ raised to the power of $\frac{3}{4}$ , multiplied by some constant.
Let's call that constant $k$ .

STEP 4

So, our function is $B = k \cdot M^{\frac{3}{4}}$ .
This captures the idea that as $M$ increases, $B$ also increases, but not at the same rate.
It's raised to the power of three-quarters!

STEP 5

The elephant's mass is 4.9 metric tons.
Since 1 metric ton = 1,000,000 grams, the elephant's mass in grams is $4.9 \cdot 1{,}000{,}000 = 4{,}900{,}000$ grams.

STEP 6

Plugging this mass into our function, we get $B_{\text{elephant}} = k \cdot (4{,}900{,}000)^{\frac{3}{4}}$ .
We don't know $k$ , but don't worry, it will disappear when we compare the rates.

STEP 7

The mouse's mass is 25 grams.

STEP 8

Plugging this into our function, we get $B_{\text{mouse}} = k \cdot (25)^{\frac{3}{4}}$ .

STEP 9

We want to know how many times greater the elephant's metabolic rate is than the mouse's.
This means we need to divide the elephant's rate by the mouse's rate: $\frac{B_{\text{elephant}}}{B_{\text{mouse}}} = \frac{k \cdot (4{,}900{,}000)^{\frac{3}{4}}}{k \cdot (25)^{\frac{3}{4}}}$

STEP 10

Notice that the $k$ values divide to one, so we're left with: $\frac{(4{,}900{,}000)^{\frac{3}{4}}}{(25)^{\frac{3}{4}}}$

STEP 11

We can rewrite this as: $\left( \frac{4{,}900{,}000}{25} \right)^{\frac{3}{4}} = (196{,}000)^{\frac{3}{4}}$

STEP 12

Now, we calculate the final result: $(196{,}000)^{\frac{3}{4}} \approx 1{,}825.74 \approx 1{,}826$

STEP 13

$B = k \cdot M^{\frac{3}{4}}$ .
The elephant's metabolic rate is approximately 1,826 times greater than the mouse's.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Write the function2. Calculate the elephant's metabolic rate3. Calculate the mouse's metabolic rate4. Compare the metabolic rates

STEP 3

"**BBB is approximately proportional to the three-quarters power of MMM**" means that BBB is almost equal to MMM raised to the power of 34\frac{3}{4}43​, multiplied by some constant.Let's call that constant kkk.

STEP 4

So, our function is B=k⋅M34B = k \cdot M^{\frac{3}{4}}B=k⋅M43​.This captures the idea that as MMM increases, BBB also increases, but not at the same rate.It's raised to the power of three-quarters!

STEP 5

The elephant's mass is **4.9 metric tons**.Since **1 metric ton = 1,000,000 grams**, the elephant's mass in grams is 4.9⋅1,000,000=4,900,0004.9 \cdot 1{,}000{,}000 = 4{,}900{,}0004.9⋅1,000,000=4,900,000 grams.

STEP 6

Plugging this mass into our function, we get Belephant=k⋅(4,900,000)34B_{\text{elephant}} = k \cdot (4{,}900{,}000)^{\frac{3}{4}}Belephant​=k⋅(4,900,000)43​.We don't know kkk, but don't worry, it will disappear when we compare the rates.

STEP 7

The mouse's mass is **25 grams**.

STEP 8

Plugging this into our function, we get Bmouse=k⋅(25)34B_{\text{mouse}} = k \cdot (25)^{\frac{3}{4}}Bmouse​=k⋅(25)43​.

STEP 9

STEP 10

Notice that the kkk values divide to one, so we're left with: (4,900,000)34(25)34 \frac{(4{,}900{,}000)^{\frac{3}{4}}}{(25)^{\frac{3}{4}}} (25)43​(4,900,000)43​​

STEP 11

We can rewrite this as: (4,900,00025)34=(196,000)34 \left( \frac{4{,}900{,}000}{25} \right)^{\frac{3}{4}} = (196{,}000)^{\frac{3}{4}} (254,900,000​)43​=(196,000)43​

STEP 12

Now, we calculate the final result: (196,000)34≈1,825.74≈1,826 (196{,}000)^{\frac{3}{4}} \approx 1{,}825.74 \approx 1{,}826 (196,000)43​≈1,825.74≈1,826

STEP 13

B=k⋅M34B = k \cdot M^{\frac{3}{4}}B=k⋅M43​.The elephant's metabolic rate is approximately **1,826** times greater than the mouse's.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Write the function
2. Calculate the elephant's metabolic rate
3. Calculate the mouse's metabolic rate
4. Compare the metabolic rates

" $B$ is approximately proportional to the three-quarters power of $M$ " means that $B$ is almost equal to $M$ raised to the power of $\frac{3}{4}$ , multiplied by some constant.
Let's call that constant $k$ .

So, our function is $B = k \cdot M^{\frac{3}{4}}$ .
This captures the idea that as $M$ increases, $B$ also increases, but not at the same rate.
It's raised to the power of three-quarters!

The elephant's mass is 4.9 metric tons.
Since 1 metric ton = 1,000,000 grams, the elephant's mass in grams is $4.9 \cdot 1{,}000{,}000 = 4{,}900{,}000$ grams.

Plugging this mass into our function, we get $B_{\text{elephant}} = k \cdot (4{,}900{,}000)^{\frac{3}{4}}$ .
We don't know $k$ , but don't worry, it will disappear when we compare the rates.

The mouse's mass is 25 grams.

Plugging this into our function, we get $B_{\text{mouse}} = k \cdot (25)^{\frac{3}{4}}$ .

Notice that the $k$ values divide to one, so we're left with: $\frac{(4{,}900{,}000)^{\frac{3}{4}}}{(25)^{\frac{3}{4}}}$

We can rewrite this as: $\left( \frac{4{,}900{,}000}{25} \right)^{\frac{3}{4}} = (196{,}000)^{\frac{3}{4}}$

Now, we calculate the final result: $(196{,}000)^{\frac{3}{4}} \approx 1{,}825.74 \approx 1{,}826$

$B = k \cdot M^{\frac{3}{4}}$ .
The elephant's metabolic rate is approximately 1,826 times greater than the mouse's.