Question

I have absolutely no idea what to do here

Answer 1

The circumference of the base changing when the radius is 8 inches is
`12\pi \text{ inches/min}`. The area of the base changing at rate
`96\pi \text{ square inches/min}` when the radius is 8 inches. The volume of the sand cone is changing at rate of
`20\pi \text{ cubic feet/min}`. The string of the kite is being let out at a rate of 3 ft/sec

For problem 11:

Given:

The radius of the base (r) is twice its height (h), i.e., r = 2h.

- The radius of the base is increasing at the rate of 6 inches per minute,
`\((dr)/(dt) = 6\)` inches/min.

a) What rate is the circumference of the base changing when the radius is 8 inches?

The circumference of the base (C) of the cone is given by
`\(C = 2\pi r\)`. We are asked to find
`\((dC)/(dt)\)` when r = 8.

`\[ C = 2\pi r \]`

`\[ (dC)/(dt) = 2\pi (dr)/(dt) \]`

Substitute the given values:

`\[ (dC)/(dt) = 2\pi \cdot 6 \]`

`\[ (dC)/(dt) = 12\pi \text{ inches/min} \]`

b) What rate is the area of the base changing when the radius is 8 inches?

The area of the base (A) of the cone is given by
`\(A = \pi r^2\)`. We are asked to find
`\((dA)/(dt)\)` when r = 8.

`\[ A = \pi r^2 \]`

`\[ (dA)/(dt) = 2\pi r (dr)/(dt) \]`

Substitute the given values:

`\[ (dA)/(dt) = 2\pi \cdot 8 \cdot 6 \]`

`\[ (dA)/(dt) = 96\pi \text{ square inches/min} \]`

c) What rate is the volume of the sand cone changing when its height is 5 feet?

The volume (V) of the cone is given by
`\(V = (1)/(3)\pi r^2h\)`. We are asked to find
`\((dV)/(dt)\) when \(h = 5\).`

`\[ V = (1)/(3)\pi r^2h \]`

`\[ (dV)/(dt) = (1)/(3)\pi (2h) (dr)/(dt) \]`

Substitute the given values:

`\[ (dV)/(dt) = (1)/(3)\pi (2 \cdot 5) \cdot 6 \]`

`\[ (dV)/(dt) = 20\pi \text{ cubic feet/min} \]`

For problem 12:

Given:

- Height of the kite (h) is 40 ft.

- The string is being let out horizontally at a rate of 3 ft/sec,
`\((dx)/(dt) = 3\)` ft/sec.

- The length of the string (s) released is 50 ft.

We want to find
`\((ds)/(dt)\)` when s = 50.

Using the Pythagorean theorem for right triangles
`(\(s^2 = h^2 + x^2\))`, we can differentiate both sides with respect to time (t):

`\[ 2s (ds)/(dt) = 2h (dh)/(dt) + 2x (dx)/(dt) \]`

Now, substitute the given values and solve for
`\((ds)/(dt)\)`:

`\[ 2 \cdot 50 \cdot (ds)/(dt) = 2 \cdot 40 \cdot 0 + 2 \cdot 3 \cdot 50 \]`

`\[ 100 (ds)/(dt) = 300 \]`

`\[ (ds)/(dt) = 3 \text{ ft/sec} \]`

So, at the given moment, the string is being let out at a rate of 3 ft/sec.