Tangent lines and differentials

[plusC]

As these questions are coming straight off my chapter review assignment and my test is coming up next week, any help with the concepts, procedures, and answers to any or (if you're feeling long winded!) all of these problems would be greatly appreciated. I understand the basics, but these are the ones that left me stumped.

1. An equation for a tangent line to the graph of y=arctan(x/3) at the origin is:

2. A solution to the equation dx/dy+2xy=0 that contains the point (0,e) is
a) y= e^(1-x^2)
b) y=e^(1+x^2)
c) y=e^(1-x)
d) y=e^(1+x)
c) y=e^(x^2)

3. If g1(x)=2g(x) and g(-1)=1, then g(x) is:
a) e^(2x)
b) e^(-x)
c) e^(x+1)
d) e^(2x+2)
e) e^(2x-2) note:
"g1(x)" is meant to be read as "g prime of x"

4. Show that if C is a constant, y=x-1+Ce^(-x) is a solution to the differential equation (dy/dx)=x-y

5. The point (1,9) lies on the graph of an equation y=f(x) for which (dy/dx)=y^(1/2)4x where x is greater than or equal to 0 and y is greater than or equal to 0. When x=0 the value of y is a) 6
b) 4
c) 2
d) 2^(1/2) or the square root of 2
e) 0

Thank you! I realize 5 is a lot of problems to post, but the assignment was huge and I am having even huger troubles trying to figure this stuff out.

 

[skeeter]

As these questions are coming straight off my chapter review assignment and my test is coming up next week, any help with the concepts, procedures, and answers to any or (if you're feeling long winded!) all of these problems would be greatly appreciated. I understand the basics, but these are the ones that left me stumped.

1. An equation for a tangent line to the graph of y=arctan(x/3) at the origin is:

for a tangent line equation, you need two things ... a point and a slope.
you have the point (0,0) ... the slope is determined by the value of dy/dx at x = 0 ... so find the derivative of y = arctan(x/3) and evaluate y'(0).
finally, use the point-slope form for a linear equation ...
y - y₀ = m(x - x₀)

2. A solution to the equation dx/dy+2xy=0 that contains the point (0,e) is
a) y= e^(1-x^2)
b) y=e^(1+x^2)
c) y=e^(1-x)
d) y=e^(1+x)
c) y=e^(x^2)

you sure it's not dy/dx + 2xy = 0 ... ?

3. If g1(x)=2g(x) and g(-1)=1, then g(x) is:
a) e^(2x)
b) e^(-x)
c) e^(x+1)
d) e^(2x+2)
e) e^(2x-2) note:
"g1(x)" is meant to be read as "g prime of x"

g'(x) = 2g(x)
g'(x)/g(x) = 2
integrate ...
ln|g(x)| = 2x + C
g(x) = e^2x+C
g(-1) = 1 = e^-2+C

so ... what is C?

4. Show that if C is a constant, y=x-1+Ce^(-x) is a solution to the differential equation (dy/dx)=x-y

y = x - 1 + Ce^-x
dy/dx = 1 - Ce^-x

so ... what is x - y?

5. The point (1,9) lies on the graph of an equation y=f(x) for which (dy/dx)=y^(1/2)4x where x is greater than or equal to 0 and y is greater than or equal to 0. When x=0 the value of y is a) 6
b) 4
c) 2
d) 2^(1/2) or the square root of 2
e) 0

dy/dx = y^1/2*4x
dy/y^1/2 = 4x dx
integrate ...
2y^1/2 = 2x² + C
now ... use your initial condition to find C and answer the question

Thank you! I realize 5 is a lot of problems to post, but the assignment was huge and I am having even huger troubles trying to figure this stuff out.

 

[plusC]

you sure it's not dy/dx + 2xy = 0...?

Hah! Yep, sorry, minor brain spasm there. But still, I can't seem to seperate the variables and I'm not sure of any other means by which I could even begin to approach it.

 

[skeeter]

you sure it's not dy/dx + 2xy = 0...?

Hah! Yep, sorry, minor brain spasm there. But still, I can't seem to seperate the variables and I'm not sure of any other means by which I could even begin to approach it.

dy/dx + 2xy = 0

dy/dx = -2xy

dy/y = -2x dx

can you finish now?

 

[plusC]

dy/dx + 2xy = 0

dy/dx = -2xy

dy/y = -2x dx

can you finish now?

Wow, I feel ridiculous, but I ended up being able to finish it. Thank you so much for your help; I should be able to ace this test next week thanks to you