Please help. I have the answers to these problems but I'd like to know how to solve it step by step.

1. Air is being pumped into a spherical balloon at the rate of 1.754 cubic centimeters per second. The balloon maintains a spherical shape throughout. How fast is the radius of the balloon changing when the diameter is 2.736 centimeters in length?

2. A wall stands upright and perpendicular from the flat ground. A 25-footlong ladder leans against the wall. The bottom of the ladder is moved away from the building horizontally (along a line perpendicular to the wall) at a rate of 30 inches per minutes (until the ladder lies flat on the ground). How fast is the top of the ladder sliding down the building when the top of the ladder is 10 feet above the ground?

3. A ladybug crawls out of a small hole in a large wall and crawls to the right at a rate of 3 inches per minute. Forty-five seconds later, a tick crawls out of the hole and crawls up at a rate of 2 inches per minute. how fast is the distance between the ladybug and the tick changing four minutes after the ladybug crawls out of the hole?

#1

I think you have a typo here. Since the r = d/2, the radius always grows half as fast as the diameter.

#2
If the base is x from the wall, and the top is y in height, then you have

x^2+y^2 = 25^2
so,
2x dx/dt + 2y dy/dt = 0
or, getting rid of the 2's,

x dx/dt + y dy/dt = 0
when y=10, x=√525 = 5√21
So, now you know x,y, and dx/dt.

#3
At time t minutes,
the ladybug has gone 3t inches
the tick has gone 2(t-3/4)

So, the distance z between them is

z^2 = (3t)^2 + (2(t-3/4))^2
z^2 = 1/4 (52t^2 - 24t + 9)
so, as in the problem above,

2z dz/dt = 1/4 (104t-24) = 26t-6
at t=4, z = √(12^2 + 6.5^2) = 13.65

So, now you know z, so figure dz/dt at t=4

i still don't understand how you did all of them. can you please elaborate? the answers for #2 is supposed to be 5.728 ft/min and #3 is 3.59in/min

Sure! I can help you with these problems one by one and explain the steps to solve them.

1. In this problem, we are given the rate at which air is pumped into a spherical balloon and we need to find how fast the radius of the balloon is changing when the diameter is a certain length. Here's how you can solve it step by step:

Let's denote the radius of the balloon as r and the time as t. We are given that the rate of change of volume (dV/dt) is 1.754 cm³/s. We need to find dr/dt when the diameter (2r) is 2.736 cm.

The volume of a sphere can be calculated using the formula V = (4/3)πr³. Therefore, the rate of change of volume can be expressed as dV/dt = (4/3)π(3r²)(dr/dt) = 4πr²(dr/dt).

We know that dr/dt = (dV/dt) / (4πr²). Substituting the given values, we get dr/dt = 1.754 / (4π(1.368)²) ≈ 0.101 cm/s.

Therefore, the radius of the balloon is changing at a rate of approximately 0.101 cm/s when the diameter is 2.736 cm.

2. In this problem, we have a ladder leaning against a wall and we need to find how fast the top of the ladder is sliding down the building at a certain point. Let's break it down step by step:

Let's denote the distance between the bottom of the ladder and the wall as x and the distance between the top of the ladder and the ground as y. We are given that dx/dt = 30 in/min (rate of change of x) and we need to find dy/dt when y = 10 ft.

Using the Pythagorean theorem, we have x² + y² = 25².

Differentiating both sides of the equation with respect to time t, we get 2x(dx/dt) + 2y(dy/dt) = 0.

Substituting the given values, we have 2(25)(30) + 2(10)(dy/dt) = 0.

Solving for dy/dt, we get dy/dt = (-2)(25)(30) / (2)(10) = -750 in/min.

Therefore, the top of the ladder is sliding down the building at a rate of 750 in/min when it is 10 ft above the ground.

3. In this problem, we have a ladybug and a tick crawling out of a hole and we need to find how fast the distance between them is changing at a certain point in time. Let's go through the steps:

Let's denote the distance traveled by the ladybug as x and the time as t. We are given that dx/dt = 3 in/min (rate of change of x) and we need to find d(distance)/dt when t = 4 minutes.

The distance between the ladybug and the tick can be calculated using the Pythagorean theorem: distance = √(x² + (x + 45/60 × 2)²).

Differentiating both sides of the equation with respect to time t, we get d(distance)/dt = (1/2)((x² + (x + 3/4)²)^(-1/2)(2x(dx/dt) + 2(x + 3/4)(dx/dt)).

Substituting the given values, we have d(distance)/dt = (1/2)((x² + (x + 3/4)²)^(-1/2)(2x(3) + 2(x + 3/4)(3)).

Evaluating d(distance)/dt at t = 4 minutes, we can substitute x = (3)(4) = 12 into the formula.

Therefore, the distance between the ladybug and the tick is changing at a certain rate which can be calculated using the above expression.

I hope this helps! Let me know if you have any further questions.