Math / Algebra

QuestionPractice Another Unclogging Arteries Research done in the 1930s by the French physiologist Jean Poiseuille showed that the resistance $R$ of a blood vessel of length / and radius $r$ is $R=\frac{k l}{r^{4}}$ , where $k$ is a constant. Suppose a dose of the drug TPA increases $r$ by $8 \%$ . How will this affect the resistance $R$ ? Assume that $l$ is constant. $\frac{d R}{R}=$

Studdy Solution

STEP 1

What is this asking? If the radius of a blood vessel increases, how much does the resistance of the blood vessel decrease? Watch out! The radius is increasing, so we expect the resistance to decrease.
Don't mix up a decrease with an increase!

STEP 2

1. Set up the equation
2. Calculate the new radius
3. Calculate the new resistance
4. Calculate the percent change in resistance

STEP 3

We're given the formula $R = \frac{kl}{r^4}$ , where $k$ is a constant and $l$ is also constant.
This tells us how the resistance $R$ relates to the radius $r$ .

STEP 4

The problem says the drug increases $r$ by 8%.
This means the new radius, let's call it $r'$ , is the original radius $r$ plus 8% of $r$ .

STEP 5

Mathematically, we can write this as: $r' = r + 0.08r = 1.08r$ So the new radius $r'$ is 1.08 times the original radius.

STEP 6

Now, let's find the new resistance, which we'll call $R'$ .
We can use the same formula, but with the new radius $r'$ : $R' = \frac{kl}{(r')^4}$

STEP 7

We know $r' = 1.08r$ , so we can substitute that in: $R' = \frac{kl}{(1.08r)^4} = \frac{kl}{1.08^4 r^4}$

STEP 8

We can rewrite this as: $R' = \frac{1}{1.08^4} \cdot \frac{kl}{r^4}$ Notice that $\frac{kl}{r^4}$ is just the original resistance, $R$ !
So, $R' = \frac{1}{1.08^4} R$ $R' \approx \frac{1}{1.3605} R \approx 0.735 R$

STEP 9

The percent change is calculated as: $\frac{\textbf{New Value} - \textbf{Old Value}}{\textbf{Old Value}} \cdot 100\%$

STEP 10

In our case, the new value is $R'$ and the old value is $R$ : $\frac{R' - R}{R} \cdot 100\%$

STEP 11

We know $R' \approx 0.735R$ , so let's plug that in: $\frac{0.735R - R}{R} \cdot 100\% = \frac{-0.265R}{R} \cdot 100\% = -0.265 \cdot 100\% = -26.5\%$

STEP 12

The resistance $R$ decreases by approximately 26.5%.

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Studdy Solution

STEP 1

What is this asking? If the radius of a blood vessel increases, how much does the resistance of the blood vessel decrease? Watch out! The radius is increasing, so we expect the resistance to decrease.
Don't mix up a decrease with an increase!

STEP 2

1. Set up the equation
2. Calculate the new radius
3. Calculate the new resistance
4. Calculate the percent change in resistance

STEP 3

We're given the formula $R = \frac{kl}{r^4}$ , where $k$ is a constant and $l$ is also constant.
This tells us how the resistance $R$ relates to the radius $r$ .

STEP 4

The problem says the drug increases $r$ by 8%.
This means the new radius, let's call it $r'$ , is the original radius $r$ plus 8% of $r$ .

STEP 5

Mathematically, we can write this as: $r' = r + 0.08r = 1.08r$ So the new radius $r'$ is 1.08 times the original radius.

STEP 6

Now, let's find the new resistance, which we'll call $R'$ .
We can use the same formula, but with the new radius $r'$ : $R' = \frac{kl}{(r')^4}$

STEP 7

We know $r' = 1.08r$ , so we can substitute that in: $R' = \frac{kl}{(1.08r)^4} = \frac{kl}{1.08^4 r^4}$

STEP 8

We can rewrite this as: $R' = \frac{1}{1.08^4} \cdot \frac{kl}{r^4}$ Notice that $\frac{kl}{r^4}$ is just the original resistance, $R$ !
So, $R' = \frac{1}{1.08^4} R$ $R' \approx \frac{1}{1.3605} R \approx 0.735 R$

STEP 9

The percent change is calculated as: $\frac{\textbf{New Value} - \textbf{Old Value}}{\textbf{Old Value}} \cdot 100\%$

STEP 10

In our case, the new value is $R'$ and the old value is $R$ : $\frac{R' - R}{R} \cdot 100\%$

STEP 11

We know $R' \approx 0.735R$ , so let's plug that in: $\frac{0.735R - R}{R} \cdot 100\% = \frac{-0.265R}{R} \cdot 100\% = -0.265 \cdot 100\% = -26.5\%$

STEP 12

The resistance $R$ decreases by approximately 26.5%.

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Studdy Solution

STEP 1

What is this asking? If the radius of a blood vessel increases, how much does the resistance of the blood vessel decrease? Watch out! The radius is increasing, so we expect the resistance to *decrease*.Don't mix up a decrease with an increase!

STEP 2

1. Set up the equation2. Calculate the new radius3. Calculate the new resistance4. Calculate the percent change in resistance

STEP 3

We're given the formula R=klr4R = \frac{kl}{r^4}R=r4kl​, where kkk is a constant and lll is also constant.This tells us how the resistance RRR relates to the radius rrr.

STEP 4

The problem says the drug increases rrr by **8%**.This means the new radius, let's call it r′r'r′, is the **original radius** rrr plus **8% of** rrr.

STEP 5

Mathematically, we can write this as: r′=r+0.08r=1.08rr' = r + 0.08r = 1.08rr′=r+0.08r=1.08r So the new radius r′r'r′ is **1.08 times** the original radius.

STEP 6

Now, let's find the new resistance, which we'll call R′R'R′.We can use the same formula, but with the new radius r′r'r′: R′=kl(r′)4R' = \frac{kl}{(r')^4}R′=(r′)4kl​

STEP 7

We know r′=1.08rr' = 1.08rr′=1.08r, so we can substitute that in: R′=kl(1.08r)4=kl1.084r4R' = \frac{kl}{(1.08r)^4} = \frac{kl}{1.08^4 r^4}R′=(1.08r)4kl​=1.084r4kl​

STEP 8

STEP 9

The percent change is calculated as: New Value−Old ValueOld Value⋅100%\frac{\textbf{New Value} - \textbf{Old Value}}{\textbf{Old Value}} \cdot 100\%Old ValueNew Value−Old Value​⋅100%

STEP 10

In our case, the **new value** is R′R'R′ and the **old value** is RRR: R′−RR⋅100%\frac{R' - R}{R} \cdot 100\%RR′−R​⋅100%

STEP 11

We know R′≈0.735RR' \approx 0.735RR′≈0.735R, so let's plug that in: 0.735R−RR⋅100%=−0.265RR⋅100%=−0.265⋅100%=−26.5%\frac{0.735R - R}{R} \cdot 100\% = \frac{-0.265R}{R} \cdot 100\% = -0.265 \cdot 100\% = -26.5\%R0.735R−R​⋅100%=R−0.265R​⋅100%=−0.265⋅100%=−26.5%

STEP 12

The resistance RRR decreases by approximately **26.5%**.

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Start learning now

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

What is this asking? If the radius of a blood vessel increases, how much does the resistance of the blood vessel decrease? Watch out! The radius is increasing, so we expect the resistance to decrease.
Don't mix up a decrease with an increase!

1. Set up the equation
2. Calculate the new radius
3. Calculate the new resistance
4. Calculate the percent change in resistance

We're given the formula $R = \frac{kl}{r^4}$ , where $k$ is a constant and $l$ is also constant.
This tells us how the resistance $R$ relates to the radius $r$ .

The problem says the drug increases $r$ by 8%.
This means the new radius, let's call it $r'$ , is the original radius $r$ plus 8% of $r$ .

Mathematically, we can write this as: $r' = r + 0.08r = 1.08r$ So the new radius $r'$ is 1.08 times the original radius.

Now, let's find the new resistance, which we'll call $R'$ .
We can use the same formula, but with the new radius $r'$ : $R' = \frac{kl}{(r')^4}$

We know $r' = 1.08r$ , so we can substitute that in: $R' = \frac{kl}{(1.08r)^4} = \frac{kl}{1.08^4 r^4}$

The percent change is calculated as: $\frac{\textbf{New Value} - \textbf{Old Value}}{\textbf{Old Value}} \cdot 100\%$

In our case, the new value is $R'$ and the old value is $R$ : $\frac{R' - R}{R} \cdot 100\%$

We know $R' \approx 0.735R$ , so let's plug that in: $\frac{0.735R - R}{R} \cdot 100\% = \frac{-0.265R}{R} \cdot 100\% = -0.265 \cdot 100\% = -26.5\%$

The resistance $R$ decreases by approximately 26.5%.