Problem Find {dy/dx} at x=1 and y=4 when x^3 + y^2 = 17 .
Answer: Use Implicit Differentiation on the equation.
It becomes 3x^2 dx + 2y dy =0
and solving for {dy/dx} we get
dy/dx =- {3x^2/2y}.
Plug in the values given for the answer -3(1)^2/(2 * 4)=-3/8 .
ProblemA 10 foot ladder is sliding down a wall. When it is 6 feet up
the wall, it is
sliding down at 2 feet/second. How fast is the base sliding horizontally?
Answer: Draw a diagram where the ladder is the hypoteneuse and the walls
are the sides of a right triangle. Call the bottom length b .
Call the height on the side wall h .
The hypoteneuse has length 10 .
The Pythagorean Theorem says that b^2 + h^2 =(10)^2=100.
Take the implicit derivative with respect to time to get
2b {db/dt} + 2h {dh/dt}=0.
We are given h=6 and, by the Pythagorean Theorem,
b = sqrt{100-36}=8.
Also, we are given that {dh/dt}=-2 .
Finally, plug in the values and solve for db/dt , that is,
the rate of change
of the length of the base.
The equation is 2(8)db/dt +2(6)(-2)=0 ,
which implies that db/dt= 3/2 .
ProblemWater is being pumped into a right circular cone
at a rate of 2 gal/min.
The height of the cone is 5 ft, and the radius of the top of the cone
is 3 ft.
When the height of the water is 2 ft,
how fast is the water rising?
Answer: Use the formula V= 1/3 pi r^2 h for the volume of a right circular cone.
At any given height, the water in the cone is in the same shape
as the the right circular cone, just a different size.
Call r the radius of the cone of water, and call h its height.
Since the shapes of the two cones are the same, you can use similar triangles
to equate the ratios of their sides.
Use the right triangle with one side being the altitude line in the center of the
cone, which has length h , and
the other side a radius of the circle, which
has length r .
In particular since the cone has radius 3 and height 5 ,
the ratio for the cone of water is r/h = 3/5 , and
we get the formula r=3h/5 .
Plugging this into the equation for volume,
the volume of the cone of water at height h is given by
V= {1/5} pi h^3 .
To find the relations between the rates of change,
we take the implicit derivative
of this equation with respect to time.
We get
dV/dt = {3/5} pi h^2 dh/dt.
Finally, plug in the given information and solve for dh/dt .
We get 2= {3/5} pi (2)^2 dh/dt so
{dh/dt}= 5/{6 pi} .
ProblemUse differentials to approximate sqrt{16.1}
Answer: Use the formula for approximations
f(x + Delta x) = f(x) + f'(x) Delta x
by plugging in x=16 and
Delta x =.1 . The answer is approximately 4+ (1/8)(.1)=4.0125 .
Problem What are the critical points for finding maximums and
minimums in
the closed interval [-1,4] for the function f(x)=|x^2-2| ?
Answer: The critical points come in three types: endpoints,
stationary points, and singular points.
The endpoints are -1 and 4 .
There is only one stationary point in the domain at 0 , and
there
are two singular points but only sqrt{2} is in the domain.
The critical points are {-1, 0, sqrt{2}, 4 } .
Note the minimum value is 0 which occurs at x= sqrt{2} ,
and the maximum value is 14 which occurs at x=4 .
ProblemFarmer Joe has 100 feet of fence to make a rectangular pen next
to a side of his barn.
The barn makes one side of the rectangle and the fence
makes the other three sides.
How long should the three other sides be
to maximize the area?
Answer: Two of the other sides are perpendicular to the barn and will
have the same length.
Call the length of these sides x .
Call the length of the third side parallel to the barn y .
The sum of their
lengths is 2x+y which must equal the length of fence, 100 .
So y= 100 - 2x .
The area of the pen is A=xy
because it is rectangular.
Plugging in we get A=x(100-2x)=100x-2x^2
as a function of one variable. The domain for x is between 0 and 100 .
The stationary points are where 0=A'(x)=100-4x , that is, x=25 .
The critical points for A are 0, 25, and 100 ,
where A attains the values
0, 1250, and 0 respectively.
Thus the maximum area is attained when
the sides of the pen are 25, 50 and 25 .
Problem{Find the maximums for the following functions on the interval [1,3] :
1/x , 5x-x^2 , x^2+1 , x/{4+x^2}
Answer: The endpoints are 1 and 3 .
We need to find the stationary points in the domain
since there are no singular points for any of these functions.
The derivative of (1/x) is -1/x^2 which has no zeros in the domain.
Since 1/1 > 1/3 , the maximum is 1 at x=1 .
The derivative of 5x-x^2 is 5-2x which has a zero at 5/2 which is
in the domain.
We find the maximum by evaluating at the critical points:
5(1)-(1)^2=4 , 5(5/2)-(5/2)^2=25/4 , and 5(3)-(3)^2=6 .
The maximum
is 25/4, which is greater than 6=24/4, at x=5/2 .
The derivative of x^2+1 is 2x which has no zeros in the domain.
Since (1)^2+1<(3)^2+1 , the maximum is 10 at x=3 .
The derivative of x/{4+x^2} is {4-x^2/(4+x^2)^2}
which has a zero at x=2 .
At the critical points 1,2 and 3 we get
{1/5}, {2/8}=1/4, and {3/13} .
Since 3/13<3/12, the maximum is 1/4 at x=2 .
}
Problem{Give an example of an increasing function which is concave up.
Answer: One possibility if f(x)=x^2 on the domain of positive numbers.}
Problem{Draw a graph for the function f , given the graph of f' .}
Problem{Give the formal definition of the limit lim {x -> infty} f(x ) = L .}
Problem Give an example of a function which does not have a limit at infty .
Answer: The basic example for this is f(x)= sin x which oscillates and has no limit at infinity. }
ProblemCalculate lim {x -> infty } 5x^2/{x^2+3x+4} .
Answer: Since 5x^2/{x^2+3x+4} = 5/{1+3/x+4/x^2} , we can plug in 0 for the limit of 1/x , and get 5/1=5 as the limit.
Problem{Find the point on the parabola y=x^2 which is closest
to the point (2, {1/2}) .
Note that there was a typo on the sheet handed out.
Answer: The distance formula between points in the plane is derived
from the Pythagorean theorem.
Between (x,y) and (2, {1/2}) it is given
by
s= sqrt{(x-2)^2+(y- {1/2})^2}.
The problem asks to minimize s given that y=x^2 . Plugging in
we get a function of one variable
s(x)= sqrt{(x-2)^2+(x^2- {1/2})^2} .
Its derivative is s'(x)=[(x-2)+2x(x^2- {1/2})]/s . Setting this equal
to zero we get 2x^3 -2=0 .
This factors as 2(x-1)(x^2+x+1).
There is only one real solution for critical points at x=1 .
The minimum distance between the parabola and the point (2, {1/2})
is the distance between (1,1) and (2, {1/2}) . The closest
point on the parabola is (1,1) , with a distance of
s= sqrt{(1-2)^2+(1- {1/2})^2}= sqrt{1+1/4}= { sqrt{5}/2} .
}