urgent help in calculus needed!!?

Question

hi,

i hate to do this for i become too much dependent relying on your answers., however, please help me with these few questions. it really makes my brain twist..

1.) a child is flying a kite at a height of 40 ft, that is moving horizontally at a rate of 3ft/sec. If the string is taut, at what rate is the string being paid out when the length of the string released is 50 ft?

2.) sand is being dropped at the rate of 10m^3/min into a conical pile. If the height of the pile is always twice the base radius, at what rate is the height increasing when the pile is 8m high?

PS
please show your solutions. i think, these 2 problems are examples of the application of derivative .
in question # 2, i get 0.597m/min, am i right??

aarghh...i certainly need an answer to this..

thank you so much to those who would help..
:-)good day!

hmmm.. how about this one.

a spherical snowball is being made so that its volume is increasing at the rate of 8ft^3/min. Find the rate at which the radius is increasing when the snowball is 4ft in diameter.

should the answer be 0.239 ft/min??

Answer 1

1) A child is flying a kite at a height of 40 ft, that is moving horizontally at a rate of 3ft/sec. If the string is taut, at what rate is the string being paid out when the length of the string released is 50 ft?

Let
x = horizontal distance
40 = vertical distance
z = length of string

dx/dt = 3 ft/sec

Find dz/dt.

We have a right triangle.

z² = x² + 40² = x² + 1600
z² = x² + 1600

x² = z² - 1600
x = √(z² - 1600)

Differentiate implicitly.

z² = x² + 1600

2z(dz/dx) = 2x
z(dz/dx) = x
dz/dx = x/z

dz/dt = (dz/dx)(dx/dt) = (x/z)(3) = 3x/z
dz/dt = 3√(z² - 1600) / z

Now plug in the values.

dz/dt = 3√(z² - 1600) / z
dz/dt = 3√(50² - 1600) / 50
dz/dt = 3√(2500 - 1600) / 50
dz/dt = 3√900 / 50 = 3*30/50 = 90/50 = 9/5 ft/sec
____________________

2) Sand is being dropped at the rate of 10m³/min into a conical pile. If the height of the pile is always twice the base radius, at what rate is the height increasing when the pile is 8m high?

Let
V = volume
h = height cone
r = base radius

dV/dt = 10m³/min
h = 2r

r = h/2

Find dh/dt when h = 8 m.

V = πr²h/3 = π(h/2)²h/3 = πh³/12

Take the derivative.

dV/dh = πh²/4

dh/dt = (dV/dt)/(dV/dh) = 10/(πh²/4) = 40/(πh²)

Plug in the value h = 8.

dh/dt = 40/(πh²) = 40/(64π) = 5/(8π) = 0.1989436 m/min

Answer 2

♥ vector v=3ft/s has 2 components: t is tangent to the kite, n is normal to the kite;
♠ the slope of the string is p=asin(40/50), therefore
n= |v|*cos(p) = |v|*√(1-(sin(p))^2 = 3*√(1-(4/5)^2) = 9/5 ft/s;
♣ volume of conic v=pi*r^2*h/3, where r=h/2; thus v=(2/3)*pi*(h/2)^3;
dv/dt =10 = (dh/dt)* (2/3)*3*pi*(h/2)^2 /2 =(dh/dt)* pi*(h/2)^2, hence
dh/dt= 10/(pi*(h/2)^2) =10/(pi*(8/2)^2)= 0.200 m/min;

Answer 3

Yes, its about the applications of the derivatives (rate of change)

I also hate these kinds of problem, sorry, can't help...☺

Answer 4

This is what I got for the Snowball problem...
(dv/dt)=8 (dr/dt)=? when d=4
dv/dt = (4/3)(π)(3r^2)(dr/dt)
8=(4/3)(π)(3)(4)(dr/dt)
dr/dt=1/2π ft/min