Mathematics: Post your doubts here!

[ffaadyy]

A2 trig eg no.5

In the first part, we need to find out the values of 'R' and 'α'. We are given the equation 'cos θ + (√3) sin θ' and using this we can easily calculate the values of 'R' and 'a'. To find 'R', we'll use the formula √[(a)^2 + (b)^2] and to find 'α', we'll use the formula tan α = (b)/(a). Doing it step by step, here's how it's done:

Finding 'R':

R=√[(a)^2 + (b)^2]
R=√[(√3)^2 + (1)^2]
R=√[3+1]
R=√4
R=2

Finding 'α':

tan α = (b)/(a)
tan α = (√3)/(1)
α = tan^-1 (√3)
α = 1.05 rad.

We can now insert the values of 'R' and 'α' in the given equation 'R cos(θ − α)' so that it becomes '2 cos (θ - 1.05)'.

Now the second part of the question consists of integration. The attached file will tell you how it's done:

 

[BeeBee]

In the first part, we need to find out the values of 'R' and 'α'. We are given the equation 'cos θ + (√3) sin θ' and using this we can easily calculate the values of 'R' and 'a'. To find 'R', we'll use the formula √[(a)^2 + (b)^2] and to find 'α', we'll use the formula tan α = (b)/(a). Doing it step by step, here's how it's done:

Finding 'R':

R=√[(a)^2 + (b)^2]
R=√[(√3)^2 + (1)^2]
R=√[3+1]
R=√4
R=2

Finding 'α':

tan α = (b)/(a)
tan α = (√3)/(1)
α = tan^-1 (√3)
α = 1.05 rad.

We can now insert the values of 'R' and 'α' in the given equation 'R cos(θ − α)' so that it becomes '2 cos (θ - 1.05)'.

Now the second part of the question consists of integration. The attached file will tell you how it's done:

View attachment 3584

thanks much

 

[ffaadyy]

thanks much

you're welcome.

 

[Scafalon40]

given that (d^2)y/dx^2=2x+1 (i.e the second derivative) and that dy/dx=y=3 when x=-1, find y in terms of x.
Ans y=(x^3)/3 +(x^2)/2 +3x +35/6

 

[rz123]

integrate double derivative to find the gradient function : 2x+1 whcih will be 2x^2/2 + x = dy/dx

integrate now to find the equation: which will be 2x^3/6 + x^2/2 +c =y

now put x value -1 into this equation and y = 3 as well to find the constant c.

c will come 17/6. so the final answe is x^3/3 + x^2/2 + 17/6 = y

my answer doesn't matches. don't know why? :/

 

[Scafalon40]

integrate double derivative to find the gradient function : 2x+1 whcih will be 2x^2/2 + x = dy/dx

integrate now to find the equation: which will be 2x^3/6 + x^2/2 +c =y

now put x value -1 into this equation and y = 3 as well to find the constant c.

c will come 17/6. so the final answe is x^3/3 + x^2/2 + 17/6 = y

my answer doesn't matches. don't know why? :/

could you tell me what symbols are used when integrating the double derivative

 

[Nibz]

∫ d^2x / dy^2 dx or ∫ f '' (x)

 

[ffaadyy]

given that (d^2)y/dx^2=2x+1 (i.e the second derivative) and that dy/dx=y=3 when x=-1, find y in terms of x.
Ans y=(x^3)/3 +(x^2)/2 +3x +35/6

integrate double derivative to find the gradient function : 2x+1 whcih will be 2x^2/2 + x = dy/dx

integrate now to find the equation: which will be 2x^3/6 + x^2/2 +c =y

now put x value -1 into this equation and y = 3 as well to find the constant c.

c will come 17/6. so the final answe is x^3/3 + x^2/2 + 17/6 = y

my answer doesn't matches. don't know why? :/

First of all, you'll integrate (d^2)y / (dx^2) = 2x + 1 to obtain the gradient equation. The gradient equation will come out to be (dy) / (dx) = (x^2) + x + c. To find the value 'c', you'll substitute dy/dx=3 and x=-1 in this equation. The value of 'c' comes out to be 3. Now you have the expression (dy) / (dx) = (x^2) + x + 3. Next, you'll integrate this equation to obtain the line equation which'll be in terms of 'y'. The line equation obtained after integrating the gradient function is y = [(x^3)/3] + [(x^2)/2] + 3x + c. Again you'll insert the values of y=3 and x=-1 in this equation to find the value of 'c' which comes out to be '35/6'. Putting this value of 'c' into the equation gives you the answer [(x^3)/3] + [(x^2)/2] + 3x + (35/6).

 

[Scafalon40]

for a single derivative we integrate like this:
dy/dx=3x+1
dy=3x+1 dx
Integrating both sides:
∫ dy=∫ 3x+1 dx
y=(3/2)x^2+x+c
but what symbols do we use for the double derivative after this step
∫ d^2y=∫ 3x+1 dx^2
i understand the method but am unsure of the symbols

 

[Nibz]

I just posted that. Don't you read all the posts?

 

[Scafalon40]

I just posted that. Don't you read all the posts?

you should really try to understand all the posts
let me make it obvious
WHAT NOTATION DO I USE AFTER THIS STEP:
∫ d^2y=∫ 3x+1 dx^2
jeesh, don't you bother to read the whole post

 

[Nibz]

You don't make sense. Word your post well.

 

[Scafalon40]

You don't make sense. Word your post well

you could at least try to read the "but what symbols do we use for the double derivative after this step" part
what part of "but what symbols do we use for the double derivative after this step"
don't you get?

 

[ffaadyy]

WHAT NOTATION DO I USE AFTER THIS STEP:
∫ d^2y=∫ 3x+1 dx^2
jeesh, don't you bother to read the whole post

Simply write it as (dy)/(dx) = (3x^2)/2 + x + c in the next step.

 

[Scafalon40]

Simply write it as (dy)/(dx) = (3x^2)/2 + x + c in the next step.

thanks

 

[cHeStEr]

Hey people!
Can y'all help me with this --->
1)Find the coordinates of the point of intersection of the tangents to the graph y=x^2 at the points at which it meets the line with equation y=x+2.
2)Find the equation of the tangent to the curve with equation y=x^3+3x^2-2x-6 at the point where it crosses the y-axis.

 

[ffaadyy]

2)Find the equation of the tangent to the curve with equation y=x^3+3x^2-2x-6 at the point where it crosses the y-axis.

The point where it crosses the 'y-axis', x=0 over there. Differentiate 'y = (x^3) + (3x^2) - 2x - 6' with respect to 'x' to obtain the gradient function. The gradient function is '(dy/dx) = (3x^2) + 6x - 2'. Insert 'x=0' in this equation to get a value for the gradient which is '-2' in this case. Similarly, insert the value 'x=0' in the equation 'y = (x^3) + (3x^2) - 2x - 6' to get a value for 'y', which comes out to be '-6'. Next, insert the values of 'y', 'x' and 'm' in the equation 'y - y1 = m (x - x1)' to find the equation of the tangent. The answer is 'y + 2x = -6'.

 

[cHeStEr]

Thank you.
And can someone help me with the 1st question of mine?
Thanks in advance

 

[ffaadyy]

Thank you.
And can someone help me with the 1st question of mine?
Thanks in advance

First of all, you'll simultaneously solve the curve equation and the line equation.

y = x^2
y = x + 2

x^2 = x + 2
x^2 - x - 2 = 0
x^2 - 2x +1x -2 = 0
x ( x - 2) + 1 (x - 2) =0
x = 2, x = -1

Now we have the x-coordinates of the point at which the curve meets the line. Accordingly, we'll find their y-coordinates which when calculated are y = 4 and y = 1.

So now we have two points, (2 , 4) and (-1 , 1). Next, we'll differentiate the curve equation 'y = x^2' to obtain the gradient function which is '(dy/dx) = 2x'. We'll insert the value x=2 in this equation to obtain a value of the gradient. The value of gradient is '4'. Inserting the values of y=4, x=2 and m=4 in the equation 'y - y1 = m (x - x1)', we get the first equation of the tangent which's 'y = 4x -4'. Similarly, using the other point which's (-1 , 1), we'll insert the value x = -1 in the gradient function to again obtain a value of the gradient which, for this point, comes out to be 'm = -2'. Again, we'll substitute the values of x = -1, y = 1 and m = -2 in the equation 'y - y1 = m (x - x1)' to obtain a second equation of the tangent which's 'y = -2x - 1'. Lastly, we'll solve both the equations of the tangents simultaneously to find their points of intersection.

-2x-1 = 4x-4
3=6x
.5=x

y=4x-4
y=4(.5)-4
y=-2

Therefore, the coordinates of the point of intersection of the tangents to the graph y=x^2 are ( .5 , -2 ).

 

[cHeStEr]

Thank you once again
I appreciate it (Y)