Fall '24 Midterm 2 Review

Question 1

The Volume of a sphere is given by the formula ‘ V = 4/3πr³’, with V = volume and r = radius. If a spherical balloon is being inflated at a rate of 160π cubic meters per minute. How fast is its radius changing at the instant when the radius is 2 meters long?

Answer 1

So first I would figure out what values and formula is given.

V = 4/3πr³
dV/dt = 160π
R = 2 meters

Looking at the formula given, I would find the derivative of V = 4/3πr³.

V = 4/3πr³
dV/dt = 4πr² dr/dt

Now that the equation is complete from the derivative, we now can plug in our values.

160π = 4π(2)² dr/dt

After which we just solve the problem and find ‘dr/dt’

160π = 16π dr/dt

Divide both sides and you will get..

dr/dt = 10m/min.

Question 2

Find the x-values at which the function f(x) = x + 3/x has a local maximum or a local minimum.

Answer 2

So we’re given the function f(x) = x + 3/x, to find the local maximum or minimum we would have to take the first derivative from the function.

f(x) = x + 3/x
f’(x) = 1 - 3/x²

Now that we know what the first derivative is, we need to establish what our zeros or critical points are.

f’(x) = 0
1 - 3/x² = 0

So we already have a zero but we need to break down 1 - 3/x^2 = 0 to get our values.

1 - 3/x² = 0

Move the -3/x^2 over to make it a positive value.

1 = 3/x²
3 = x²

Then multiply each side by 3 to keep x^2, then square root both sides in order to get

+ sqrt( 3
- sqrt( 3

Now we need to really find the local maximums and minimums

So now we need to use the 2nd derivative

f(x) = x + 3/x
f’(x) = 1 - 3/x²
f’’(x) = 6/x³

Now we plug in our zeroes / critical points

6/(+sqrt(3))³ = 1.15 (+)
6/(-sqrt(3))³ = -1.15 (-)

Now in order to figure out what’s the local maximum and minimum, we need to consider the 2nd derivative rule as like a room swap where everything is opposite.

So – sqrt(3 would be the local maximum and sqrt(3 would be the local minimum.

Question 3

Locate the absolute extreme values of the function g(t) = ⅔ x³ - ½ x² on the closed interval [0,3].

Answer 3

We’re being dealt with the function g(t) = 2/3x³ - 1/2x² so the first thing we should do is to find the 1st derivative of the equation

g(t) = ⅔ x³ - ½ x²
g’(t) = 2x²-x

Then you need to determine the ‘zeroes’ of the function which is g(t) = 0

g’(t) = 2x²-x = 0

Factor to find the 2nd with the inside function

x(2x-1) = 0
2x-1 = ½

Now that we know the zeroes, we now need to plug these values back into the 1st function and evaluate in order to find the absolute maximum and minimum

g(0) = ⅔ (0)³ - ½ (0)² = 0
g(½) = ⅔ (½)³ - ½ (½)² = -1/24
g(3) = ⅔ (3)³ - ½ (3)² = 13.5 or 27/2

The absolute maximum is 27/2, and the absolute minimum value is -1/24.

Question 4

Find the critical point(s) of the function

g(x) = x²/x-2

Answer 4

First I would establish what I should do based on the wording, so if we were trying to find the critical point we would have to find the first derivative.

Based on how the function is formed you would have to use the quotient rule HidLo-LodHi/Lo²

((x²)(1)) - ((x-2)(2x)) / (x-2)²

2x(x-2) - x² / (x-2)²

I would then distribute the 2x in the numerator to (x-2)

2x² - 4x - x² / (x-2)²

After that you would need to combine like terms

2x² - 4x - x² / (x-2)²

X² -4x / (x-2)^2

Now you have your fractions needed to find the critical points

Numerator distribute the x and find the critical point

x² - 4x = 0
x(x-4) = 0
(x-4) = 4

Denominator find the critical point
(x-2)² = 2

So our critical points are

0, 2 and 4

Question 5

Evaluate the derivative of the function f(x) = 7 · 3ˣ - logₑ(x²) at x = 2

Answer 5

So first you need to split this functions into different sections to focus on, so we’ll focus on 7 · 3ˣ

Take the derivative of this section, but keep in mind that 7 stays in the derivative because its a constant multiplier
7 · 3ˣ
7 · 3ˣln(3)

Then take the derivative of logₑ(x²)
*keep in mind that the derivative logₑ(x²) is ln(x²) because ln is just another way of saying natural log of number or value. Then ln(x²)’s derivative is 2/x (i’ll put it into another flashcard later.)

logₑ(x²)
ln(x²)
2/x

So now we’re left with this equation, now we can consider plugging in the x value which is 2
7 · 3ˣln(3) - 2/x

7 · 3²ln(3) - 2/2

7 · 9²ln(3) - 1

Simplify and you’ll get this result
63ln3 - 1

Question 6

Find dy/dx for the equation

xy = 5y² + 2

Answer 6

First when we’re looking at what they’re asking its usually asking for implicit differentiation so now that we know that lets split it into two parts of focus, the left and right side, so we’ll start off with the left which is xy

We know for certain that we can use the product rule in the sequence, so consider LdR + RdL

xy
d/dx [xy] = d/dx(x) · y + d/dx(y) · x
y + x dy/dx

Now that we have the left side we can move onto the right side d/dx[ 5y² +2] remember that 2 is disregarded because the derivative of that is 0.

5y² + 2
10y dy/dx

So now we’re left with the equation y + x dy/dx = 10y dy/dx

y + x dy/dx = 10y dy/dx

You then move x dy/dx to the other side making it

y = 10y dy/dx - x dy/dx

After you would need to factor out the dy/dx from the 10y-x

y = (10y - x) dy/dx

Divide both sides by (10y - x) and the result will be

dy/dx = y / (10y - x)