# Thread: Basic Calc I questions (regarding derivatives, instantaneous rate of change, etc.)

1. ## Basic Calc I questions (regarding derivatives, instantaneous rate of change, etc.)

1) A particle is moving along the curve y = sqrt(x). As the particle passes through the point (1,1), its x-coordinate increases at a rate of 5 cm/sec. How fast is the distance D from the particle to the origin changing at this instant?

2) Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 23 feet high? Recall that the volume of a right circular cone with height h and base radius r is given by v=1/3pi*r^2h.

I'm very lost, and don't know what to do. I know I should draw a graph and take the derivative based off of an equation derived from the graph, but idk!

I really appreciate it

2. Originally Posted by gobbajeezalus
A particle is moving along the curve y = sqrt(x). As the particle passes through the point (1,1), its x-coordinate increases at a rate of 5 cm/sec. How fast is the distance D from the particle to the origin changing at this instant?

I'm very lost, and don't know what to do. I know I should draw a graph and take the derivative based off of an equation derived from the graph, but idk!

I really appreciate it
at any time $t$ , the particle is at the point $(x,\sqrt{x})$

distance between the particle and the origin is

$r = \sqrt{(x-0)^2 + (\sqrt{x} - 0)^2} = \sqrt{x^2 + x}$

you were given $\frac{dx}{dt}$ ... find $\frac{dr}{dt}$ when $x = 1$

3. Originally Posted by gobbajeezalus
2) Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 23 feet high? Recall that the volume of a right circular cone with height h and base radius r is given by v=1/3pi*r^2h.
given $\frac{dV}{dt} = 10 \, ft^3/min$

$V = \frac{\pi}{3}r^2 h$

since diameter = height ... $r = \frac{h}{2}$

$V = \frac{\pi}{3} \left(\frac{h}{2}\right)^2 h = \frac{\pi}{12} h^3$

take the derivative w/r to time and determine the value of $\frac{dh}{dt}$