Calc/Stat – Mylod Unit 6 Review Problems – Applications of Derivatives Find the derivatives (no calculator): Name: Write the equation of the tangent line to f (x) = x 2 + 2x ! 6 at x=-­‐3 (no calculator) d
sin x 3 = 3x 2 cos(x 3 ) dx
y = !4x ! 15 d
5
4
( 3x + 5 ) = 15(3x + 5) dx
d sin x
1
=
dx cos x cos 2 x
When f is increasing, the derivative is Positve d 4x
4x
x
4 = 4 i ln 4 i 4 i ln 4 dx
When f is decreasing, the derivative is Negative When a particle changes direction, the velocity d
cos ( x ) ! 27 x = 27 x ( ln 27 cos(x) " sin(x)) dx
Changes sign When f is concave up, the second derivative is d x 22 22x 21 ! x 22
Positive =
dx e x
ex
When f is concave down, the second derivative is Negative d
sin 3x i 3x 4 = cos 3x i 3x 4 !" 3x i12x 3 + 3x 4 i 3x i ln 3#$ dx
( )
(
(
)
)
(
)
Let x(t) = (x ! 3)5 be the position of a particle at time t, find the velocity and acceleration of the particle at time t (no calculator). v(t) = x !(t) = 5(x " 3)4 a(t) = x!!(t) = 20(x " 3)3 1990 – BC1
Show a complete sign chart as part of your response to the
following:
Let v(t) = x !(t) = cost be the velocity of a particle at time t, find the acceleration of the particle at time t. a(t) = x!!(t) = " sin(t) For 1 ! t ! 4 , when is the particle at its maximum velocity and acceleration? Maximum velocity at t = ! = 3.14 Maximum acceleration at t = 4 Sketch a graph of the following (without using a calculator. y = ( x 2 ! 4 ) 2
A particle starts at time t = 0 and moves along the x-axis so
that its position at any time t ≥ 0 is given by
x(t ) = (t −1) ( 2t − 3)
3
a) Find the velocity of the particle at any time t ≥ 0 .
x '(t) = (t ! 1)2 (8t ! 11)
b) For what values of t is the velocity of the particle less
than zero.
at y ' = 0 , x = !2,0,2 max at x = 0 , min at x = !2,2 at y" = 0 , x = ±
x"(t) = 6(t ! 1)(4t ! 5)
d) At time t = 2 is the particle speeding up or slowing
down?
! ±1.15 ( !2,0) ,( !1.15,7.11) ,(0,16) ,(1.15,7.11) ,( 2,0) c) Find the value of t when the particle is moving and the
acceleration is zero.
x"(t) = 18
3
Critical points on y are: ( !",1) # $&% 1, 118')(
x '(t) = 5
2
Speeding up! Find two numbers whose sum is 20 and whose product is as large as possible. x = y = 10
xy = 100
You have been asked to design a one-­‐liter oilcan shaped like a right circular cylinder. If r and h are measured in centimeters, then the volume of the can in cubic centimeters is, The surface area of a can is ! r 2 h = 1000 (1 liter = 1000 cm 3 ) . 2! r 2 = Area of the circular ends
SA = 2! r + 2! rh, where 2! rh = Area of the cylinder wall
2
What dimensions of the oilcan will use the least material (ignore thickness of the material used)? r = 5.42 cm, h = 10.84 cm
SA = 553.73 cm2