# 1) Let F(x)= f(x^{5}) and G(x)=(f(x))^{5} . You also know that a^{4}=13, f(a)=3,f'(a)=9, f'(a^{5})=12. Find F'(a) and G'(a). 2) Water is leaking out of an inverted conical tank at a rate of 9500.0...

1) Let F(x)= f(x^{5}) and G(x)=(f(x))^{5} . You also know that a^{4}=13, f(a)=3,f'(a)=9, f'(a^{5})=12. Find F'(a) and G'(a).

2) Water is leaking out of an inverted conical tank at a rate of 9500.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6.0 meters and the diameter at the top is 6.5 meters. If the water level is rising at a rate of 24.0 centimeters per minute when the height of the water is 3.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute. Note: Let "R" be the unknown rate at which water is being pumped in. Then you know that if V is volume of water, \frac{dV}{dt}=R-9500.0. Use geometry (similar triangles?) to find the relationship between the height of the water and the volume of the water at any given time. Recall that the volume of a cone with base radius r and height h is given by (1/3)pir^2h

3) Let A be the area of a circle with radius r. If dr/dt =2, find dA/dt when r= 2.

4) let xy= 2 and let dy/dt= 3. Find dx/dt when x= 3.

5) A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.1 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 10 cm. (Note the answer is a positive number).

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(2) We are given that the rate at which water is leaving the tank is ` `` ``(dV)/(dt)=R-9500"cc"/"min" ` where R is the unknown rate of additional water being poured into the tank.. The height H of the tank is 6m, the diameter of the top is 6.5m. The rate at which the depth of the water is increasing is `(dh)/(dt)=24"cm"/"min" ` where h is the depth of the water when h=3.5m.

The volume of a right circular cone is `V=(pi r^2h)/3 ` . ` `If the depth of the water is h and the radius of the cone containing the water is r, then by similar triangles we have `3.25/6=r/h ` . (3.25m is the radius of the circular opening of the tank, and 6m is the height of the tank.) Then we can express r as a function of h: `r=3.25/6*h ` .

Substituting we get:

`V=1/3*pi*(3.25/6h)^2*h ` or `V=(169pih^3)/1728 `

Then `(dV)/(dt)=(169pih^2)/576 (dh)/(dt) `

Now with h=3.5m or 350cm, ` (dh)/(dt)=24"cm"/"min" ` ` `

`R-9500"cc"/"min"=(169pi(350"cm")^2)/576*24"cm"/"min" `

`R-9500=(496860000pi)/576 "cc"/"min" `

`R~~2719450.9"cc"/"min" ` or `R~~2.72"m"^3/"min" `

(3) We know ` A=pir^2 ` so `(dA)/(dt)=2pir(dr)/(dt) ` .

Here we are given `(dr)/(dt)=2,r=2 ` so `(dA)/(dt)=2pi(2)(2)=8pi~~25.1 `

(4) Given `xy=2,(dy)/(dt)=3, x=2 ` :

We have `x(dy)/(dt)+y(dx)/(dt)=0 ` Substituting the known values:

`2(3)+1(dx)/(dt)=0 ` (xy=2 implies y=1 when x=2.)

`(dx)/(dt)=-6 `

(5) The volume of a sphere is `V=4/3pir^3 ` or in terms of the diameter

`V=4/3pi(d/2)^3=(4pid^3)/24=(pid^3)/6 `

Then `(dV)/(dt)=(pid^2)/2 (dd)/(dt) ` Substituting known values we get:

`(dV)/(dt)=(pi*100)/2(.1)=5pi~~15.71("cm"^3)/"min" `

To answer problem no. 1 we have to recall the formula for chain rule .

The derivative of a function like (c(x))^n is n*(c(x)^(n-1)) * c'(x) .

With this, we can get the derivative of F(x) = f(x^5) by taking the derivative of the outer function f first then using the chain rule for the inner function x^5, so F'(x) = f'(x^5)*5x^(5-1) simplifying this we will have F'(x) = 5f'(x^5)*x^4.

To get F'(a) we plug-in a for x in the derivative we got, so F'(a) = 5f'(a^5)*(a^4) and using the information given in the problem it will be 5*(12)*(13)=780.

The same with F'(a), we get G'(a) using the chain rule. However, in this case it is in the outer function that applies. G(x) = (f(x))^5 has the derivative of G'(x) = 5(f(x))^(4)*f'(x).

So, G'(a)= 5(f(a))^(4)*f'(a).

Again, referring to the given information we have G'(a)=5*(3)^(4)*9 = 3645.