MTH 161 Chapter 2 review
Let y^2(y^2-4) = x^2(x^2-5). Find an equation of the tangent line to the devil's curve at the point (0, -2).
4y^3y'-8yy' = 4x^3 - 10x y' = 4x^3 - 10x divided by 4y^3 - 8y at (0, -2) = 0/-8 = 0 write in slope intercept form
definition of differentiality
A function f is differentiable at a if dy/dx x=a exists. It is differentiable on every open interval (a,b) or (a, ∞) or (-∞. ∞) if it's differentiable at every number in the interval
derivative of cos
- sin x
derivative of csc x
-csc x cot x
derivative of cot x
-csc^2 x
dy/dx (c)
0
dy/dx (x)
1
find dy/dx by implicit differentiation: 1+x = sin(xy^2)
1 = cos(xy^2)(2xyy' + y^2) 1 = 2xyy' cos(xy') + y^2cos(xy^2) -2xyy'cos(xy^2) + 1 = y^2cos(xy^2) y'(2xycos(xy^2) = y^2cos(xy^2) y' = 1-y^2cos(xy^2) / 2xycos(xy^2)
differentiate the following with respect to t: x^2(t) + y^2(t) = z^2 (t)
2x(dx/dt)+2y(dy/dt) = 2z(dz/dt)
use implicit differentiation to find dy/dx when x^3 + y^3 = 6xy
3x^2 + 3y^2y' = 6xy' + 6y dy/dx 3y2-6x= 6y - 3x^2 dy/ dx = 6y-3x2/3y2-6x
distance formula
D= square root of (x2−x1)^2+(y2−y1)^2
how to find the equations of both lines through a specific point that are tangent to a curve
Find the derivative of the curve Find the secant line between the given point and the x and f(x) values Set these two derivatives equal to each other Solve for x Put those x values into the original equation to get the y values Put the x values into the derivative of the curve equation to find the slope at this point Write the equations using the point given in the problem
how to find an equation to the tangent line to a curve at a specific point
Find the derivative of the function Put the x value of the point into the derivative That will be the slope Write in slope intercept form
Show that the curves x^2 - y^2 = 5 and 4x^2 + 9y^2 = 72 are orthogonal
Find the points of intersection x^2-y^2 = 5 and 4x^2 + 9y^2 = 72 Rewrite one of the equations for x Put what x equals into the other equation to solve for y Once you get y, put the value for y into an equation to get x X = +/- 3 Y = +/- 2 Take the derivative of each equation Take one of the four points of intersection, put those values into each of the derivatives -⅔ and 3/2 If they are opposite reciprocals, they are orthogonal If not, they are not orthogonal
differentiality and continuity
If f is differentiable at a, then f is continuous at x = a *if you can determine the slope at a, then it is differentiable
non differentiable functions: absolute value function
Let y = x. The domain of y is (-∞, ∞). Let us find dy/dx using the definition of the derivative of a function as the limit of a difference quotient. Would have to look at the limit from the left and the limit from the right (x) = (x if x is greater than or equal to 0 and -x if x is less than zero) the limits do not equal each other
Let f(x) = √(x+4). Let P(5,3) be a point on the graph of f. What is the slope of the tangent line at P?
Point is (5,3) Tangent point is (x, √(x+4)), also known as Q Slope =( (√x+4) - 3) / (x-5) Finding the slope between the given point and the tangent point Take the limit as x approaches 5 Limit as x approaches the x value of the given point Simplify - in this case, multiply by the conjugate Put 5 into the equation as x Slope of the tangent = ⅙
find the equations of both lines through the point (2,3) that are tangent to the parabola y = x + x^2
Points are (2,3) and (a, a + a^2) Take the derivative of the parabola Y' = 1+ 2x Find the slope between the two points - slope of the secant a2+a+3/a-2 Set the slope of the secant and the derivative of the parabola equal to each other (rewrite the derivative of parabola to include a) Solve for a a = 5, a = -1 These are the x values of the points with lines that go through the given point Put these x values into the original equal to get the y values Put these x values into the tangent equation (derivative of parabola) to get the slope Write in slope intercept form, using the point that both lines go through @(-1,0), slope = -1 Y + 3 = -(x-2) @(5, 30), slope = 11 Y+3 = 11(x-2)
how to find the average velocity
Put each of the sides of the interval into the velocity function Find the average of those two velocities - do this by using the y2-y1 / x2-x1 formula
find an equation of the tangent line to the curve y = x√x that is parallel to the line y = 1+3x.
Rewrite original equation as y = x^3/2 Slope of the tangent line is 3, given to you in the line equation because your tangent line is parallel Means that the slope will be the same Take the derivative of the original function Set this equal to 3 Solve for x These are the x value(s) at which the original function has the slope of 3 Use this x value and find the y value using the original function Write in point slope form y - 8 = 3(x-4)
If a snowball melts so that its surface area decreases at a rate of 1 cm^2/min, find the rate at which the diameter decreases when the diameter is 10 cm.
S: surface area X: diameter V = 4/3πr^3 dv/dt = 4/3π(3r^2) = s Simplify S = πx^2 ds/st = π(2x) (dx/dt) Solve the problem dx/dt = 1/2xπ (ds/dt) = 1 dx/dt = 1/20π (1) dx/dt = 1/20π cm^2/min
how to determine where a normal/tangent line intersects a graph multiple times
Set the equation for f(x) equal to the equation for the normal/tangent line and solve for x You get the equation for the normal/tangent line through the simplified form of the slope intercept equation Put the x values back into the equation for f(x) to determine their y values
law of sines
Sin alpha / a = sin beta / b = sin y / c
how to show that two curves are orthogonal to each other
Take both given equations Rearrange one of them for one of the variables Put what this variable equals into the other equation to solve for the other variable (get a value) Use this value in one of the equations to get the other variable You should get the x and y values that you need Take the derivative of each equation Pick one of the points determined previously in the problem Use these x and y values in the derivatives to get a value at that point If the slopes are opposite reciprocals of each other, they are orthogonal If not, they are not orthogonal
how to find an equation of the tangent line to a curve at a specific point
Take the derivative by using implicit differentiation Put the x and y values into the derivative to find the slope at the point given Write in slope intercept form
how to find an equation of a tangent line to a curve that is parallel to another line
Take the derivative of the curve Find the slope of the line the curve is supposed to be parallel to Set the derivative of the curve and the slope of the line equal to each other Solve for x Put the x value into the equation of the original curve to find the y value Write in slope intercept form (y-___ = slope (x-___)
how to find points on a curve at which a tangent line is horizontal over an interval
Take the derivative of the curve Simplify as much as possible Set the numerator equal to 0 Solve Look for the corresponding values on the unit circle These are the x values Use the x values and put them into the original equation for the curve to find the y values
how to find the x coordinates of all points on a curve at which a tangent line is horizontal for an interval
Take the derivative of the curve Simplify by factoring, using identities if needed Factor again if identities are used Set the factors equal to zero Find the corresponding values on the unit circle to what you find
how to find the equation of the normal line to a graph at a specific point
Use the definition of derivative Simplify as much as possible, put either 0 or a into the equation to get the slope Normal line is the opposite reciprocal of the slope of the tangent line Y - ___ = slope(x-___) to determine the equation of the normal line
Find an equation of the tangent line to the curve y = (1+x)cosx at the point (0,1)
Y = (1+x)(cos x) Y' = cos x + (1+x)(-sin x) At x = 0, y' = 1, slope = 1 Equation: y = x + 1
find the points on the curve y = cos x divided by 2 + sin x at which the tangent is horizontal for 0 less than or equal to x less than 2 pi
Y' = (2+ sinx)(-sinx)-(cosx)(cosx) divided by (2+sinx)^2 Simplify -2sinx - 1(2+sinx)2 Set the derivative equal to 0 When the tangent is horizontal, this means that the slope is 0 This only occurs with horizontal tangents Set the numerator of the derivative equal to 0 Solve Sin x = -½ Find the values on the unit circle that correlate with sin x = -½ x = 7π/6, 11π/6 Put these values into the original equation to get the y values
Find the derivative of the function: y = cot(x^2) + cot^2 x
Y' = -2x csc(x^2) + -2csc(x)cot(x)
Apply the chain rule to find the derivative of y = (x^5-2x^2)^10
Y' = 10(x^5-2x^2)^9 x (5x^4 - 4x)
Find the x coordinates of all points on the curve y = sin2x - 2 sin x at which the tangent line is horizontal for 0 x < 3π
Y' = 2 cos(2x) - 2cosx Y' = 0 2(cos(2x)-cosx) (disregard the 2 on the outside from this point forward) Trigonometric identity 2cos^2x - 1 Rewrite using trigonometry identity 2cos^2x - cos x - 1 Factor (2 cos x+1)(cos x - 1) Determine which values correlate with cos x = -½, 1 on the unit circle These are the x values
chain rule
Y' = f'(g(x)) x g'(x)
Find the derivative of the function: y = x sin(square root of x)
Y' = sin square root of x+ x^1/2 2cos square root of x
normal line
a normal line at point P is a line that is perpendicular to the tangent line at P
law of cosines
a^2 = b^2 + c^2 -2bc Cos a
d/dx (c(f(x))
c dy/dx f(x) c is a constant
d/dx sin x
cos x
derivative of sin x
cos x
d/dx [f(x) + g(x)]
d/dx f(x) + d/dx g(x)
d/dx [f(x) - g(x)]
d/dx f(x) - d/dx g(x)
differentiate the following with respect to t: C(t) = 2πr(t)
dC/dt = 2π(dr/dt)
find a formula for the nth derivative of y = 1/x
d^n/dx^n of (1/x) = (-1)^nn!/xn+1
velocity
derivative of position
acceleration
derivative of velocity
how to find the total distance traveled in x amount of time
determine the position of the particle when the velocity is zero determine how far the particle moved between points add them all put
How to solve a related rates problem when there is a man, shadow, and he is moving toward something that the shadow is on, and deals with similar triangles?
draw a picture account for the fact that the man is between the spotlight and what the spotlight is projected on make a ratio for the similar triangles rearrange this equation for y take the derivative put in the values you known solve and simplify
how to solve related rates with conical tanks
draw a picture determine what is changing, these are your variables v = 1/3 (pi) r^2 h to get rid of two variables, set up a similar triangle ex. r/h = 2/6. r = 1/3 h replace what the r equals in the overall equation take the derivative put in the values you know make sure all of your numbers have the same units, if not change them simplify and solve for what you need
how to solve related rate problems with isosceles triangles?
draw a picture determine what variables you are solving for write a relationship you most likely are going to need to use similar triangles use similar triangles to come up with a proportion of b to h of the triangle you need rewrite the relationship using this method take the derivative put in what you know and solve
how to solve related rate problems with ships that don't start from the same point
draw a picture make sure you account for the fact that the ships don't leave and/or move from the same position determine the variables use the triangle equation: a^2 + b^2 = c^2 take the derivative put in the values you know if given only rate with a time, use the distance traveled over a rate of time to determine the total distance that those ships moved if they aren't given to you put in the values you have solve and simplify as much as possible
differentiate the following with respect to t: V = ⅓ πr^2h
dv/dt = π/3 (2r(dr/dt)h + r^2 (dh/dt))
differentiate the following with respect to t: X = cos beta
dx/dt = sin beta (d beta / dt)
derivative of a function f at point a - definition
f'(a) = lim as h approaches 0 of f(a+h) - f(a) divided by h f'(a) = lim as x approaches a of f(x) - f(a) divided by x - a
Let g(x) = 1/x. Find an equation of the tangent line to the graph of g at the point (2, 1/2).
f'(a) = lim as h approaches 0 of f(a+h) - f(a) divided by h put (2+h) and 2 into the function, put those values into the definition of derivative function simplify, find a common denominator cancel the -h and h put 2 into the simplified equation to find the slope slope = -1/4 use the form y - ___ = slope (x - ____) to determine the equation of the tangent line
Find all of the points on the curve x^2y^2 + xy = 2 where the slope of the tangent line is -1.
f'(x) = 2x^2yy' + 2xy^2 + xy' + y = 0 Y' = (-2xy^2 - y) / (2x^2y + x) = -1 Multiply across -y - 2xy^2 = -2x^2y - x y+2xy^2 = 2x^2y + x y(1+2xy) = x (1+2xy) (1+2xy)(y-x) = 0 Xy = -½ When you put it back into the original equation, it does not equal 2, it equals ¼ so it doesn't work Y = x Put back into original equation in the problem x^4 + x^2 = 2 x^4 + x^2 - 2 = 0 (x^2 - 1)(x^2 +2) = 0 (x^2+2) Not possible (x^2-1) X = +/- 1 Points are (1,1) and (-1,-1)
Use the definition of derivative as a limit to find the derivative of y = f(x) = √x.
f(x) = √x f'(x) = √(x+h) - √x divided by h simplify, multiply by the conjugate factor out what you can f'(x) = 1/2√x
product rule
f(x)g'(x) + g(x)f'(x)
how to find the total displacement traveled in x amount of time
find the distance traveled between t = 0 and t = x
how to find when a particle is at rest
find the zeroes of the velocity function
how to find when the acceleration is 0
find the zeroes of the velocity function
non differentiable functions: square root function
let y = √x. the domain of y is ‖0,∞ ) we saw earlier that dy/dx = 1/2√x at what point of the domain is y not differentiable? What does it mean for the graph of y? F'(0) does not exist y = √x has a vertical tangent at (0,0)
quotient rule
low d high - h d low divided by low low
definition of derivative
m = lim as x approaches a of f(x) - f(a) divided by x - a
how to solve related rate problems with triangles and angles?
make a picture determine what variables you are using write a relationship this most likely will ask you to use the law of cosines put in what you know to get the missing value you need take the derivative of the law of cosines put in the values you know change all values to the correct unit if needed simplify as much as possible
how to solve related rate problems with triangles, angles and area
make a picture determine what variables you need you are going to need to draw in the height the equation you need is A = 1/2bh in order to find h, use the sine of the angle you have and rearrange it to use h put the value for h into the equation for area simplify if needed take the derivative put in the values you have and solve for what you need
similar triangles
make a ratio between the similar sides (sides that aren't hypotenuses)
dy/dx (x^n)
nx^n-1
orthogonal
perpendicular lines to each other
how to find instantaneous velocity
put the t = ___ value into the velocity function
derivative of sec x
sec x tan x
derivative of tan x
sec^2 x
Let y = x - x². Find an equation of the normal to the graph of y at the point (1,0).
slope of the tangent line at x = 1 f'(1) = limit as h approaches 0 of f(1+h) - f(1) divided by h f'(1) = limit as h approaches 0 (1+h) - (1+h)² divided by h limit as h approaches 0 of -h - h² divided by h factor out an h from the numerator to cancel with the denominator -1-h, put 0 in for h slope = -1 normal line is opposite reciprocal so slope = 1 y - 0 = 1(x-1)
how to find points on a lemniscate where a tangent is horizontal
take the derivative if there are three things on one side, foil it out put all y' terms on one side of the equation simplify as much as you can set the numerator equal to 0 solve for a part from the original equation put this into the original equation and solve for another part once you have those two equations, rearrange one of them for one of the variables put this into the other equation to solve for one variable once you know that one variable, solve for the other variable in the other equation take square roots if needed, should be able to find the x and y values
how to find equations of both tangent lines to an ellipse that passes through a specific point
take the derivative of the equation put that derivative into an equation of the tangent equation rewrite the derivative in the equation of the tangent using a and b replace the x and y in the tangent as a and b multiply it out get the b^2 and a^2 on the same side and the b and a on the same side the a^2 and b^2 should be similar to the initial equation use the value that this equals and set it equal to the other side with just a and b solve for b put b back into the initial equation to solve for a now you have the two a values, put into equation to get the y values these are the points on the ellipse write tangent line equations do this by putting a and b values into the tangent equation with a and b (and with the x and y) simplify as much as possible, you have the tangent lines
how to solve a problem in which a particle is moving along a point
take the derivative of the equation you are given put in the numbers you are given or can figure out simplify and solve
how to find the velocity at time t
take the derivative of the position function
how to find the velocity at a specific time
take the derivative of the position function put the t value into the velocity function
how to find the equation of a tangent line to a graph at a specific point
use the definition of derivative simplify as much as possible, put either 0 or a into the equation to get the slope y - ___ = slope (x-___) to determine the equation of the tangent
let F(x) = 3x² + 1. Find f'(2).
use the definition of derivative in order to find the derivative lim as x approaches a of f(x) - f(a) divided by x - a plus in 3x²+1 in as x, put 2 into the given function and put that in as f(a) simplify as much as possible, in this case, factor the top and cancel terms with the bottom put 2 in for x, as x approaches 2 f'(2) = 12
how to solve related rate problems with walls and sliding ladders?
use x^2 + y^2 = L^2 take the derivative L^2 will become 0 because the height of the ladder doesn't change, can also apply to other sides of the triangle not changing either (they would be 0 because of the rate of change) solve for the value you don't know once you have that value, put it into the original equation to get the length you need
what is the volume of a conical tank
v = 1/3 pi (r^2) h
volume of a cylinder
v = pi r^2 h
A cylindrical tank with radius 5 m is being filled with water at a rate of 3 m^3/min. How fast is the height of the water increasing?
v: volume h: water level dh/dt = ? v=(pi)r^2h v= 25(pi) h dv/dt = 25(pi) dh/dh 3 = 25(pi) dh/dt 3/25(pi) = dh/dt
when are graphs not differentiable?
vertical tangent corner jump
At what point on the graph of y does the normal line at (1,0) cross the graph of y a second time?
want to find the points of intersection set the equation given at the start of the problem equal to the equation of the normal line x-1 = x - x² solve x = +/- 1 use the x values to find the y values one of the points will be the point you used previously in the problem the other point is the one you want, the one you haven't used yet
when is a particle slowing down?
when acceleration and velocity have different signs
when is a particle speeding up?
when acceleration and velocity have the same sign
when is a particle at rest?
when the velocity is 0
when is a particle moving in the negative direction
when the velocity is negative
when is a particle moving in the positive direction
when the velocity is positive
how to determine differentiality
would have to look at the limit from the left and the limit from the right to determine if a function is differentiable at a certain poiint
how to solve a problem with a cylindrical tank and volume
write the equation for volume take the derivative put in the values you know or can figure out simplify and solve
how to solve related rate problems with a surface area and diameter?
write the equation of surface area take the derivative put in the values you know rearrange certain parts of the problem if you can't solve after this you should be able to simplify as much as you can
A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.6 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?
x(t) = distance between man and spotlight y(t) = length of the man's shadow on the wall question = dy/dt similar triangles y/12 = 2/x rearrange - 24x^-1 dy/dt = -24/x^2 dx dt dy/dt = -24/(8)^2 times 1.6 dy/dt = -.6 m/s
At noon ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. how fast is the distance between the ships changing at 4:00?
x(t) = position of ship A y(t) = position of ship B z(t) = distance between A and B (150-x)^2 + y^2 = z^2 2(150-x)(dx/dt) + 2ydy/dt = 2z dz/dt put the values in get x and y values by using rates to determine how far the ships have gone put those values in and solve 2,150 / square root of 10100 = dz/dt
A particle is moving along the curve y = square root of x. As the particle passes through the point (4,2), its x coordinate increases at a rate of 3 cm/s. How fast is the distance from the particle to the origin changing at that instant?
x- coordinate of a point f - distance between the origin and point P df/dt = ? df/dt = 1/2(x^2 + x)^-1/2 (2x dx/dt + dx/dt) solve df/dt = 27/4 square roots of 5
find the derivative of the function: y = square root of x + square root of x + square root of x
y' = 1/2 (x + square root of x + square root of x)^-1/2 times (1 + 1/2(x)^-1/2)(1+1/2x-1/2)
