# Trigonometry-related calculus questions

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I would appreciate if someone could check my solutions for the following:

find dy/dx of y=sqrt(sec(5x))

find dy/dx of y+sin(y)=x

Thank you for looking , thank you even more if you help!

#### Jhevon

MHF Helper
I would appreciate if someone could check my solutions for the following:

find dy/dx of y=sqrt(sec(5x))

find dy/dx of y+sin(y)=x

Thank you for looking , thank you even more if you help!
the first is incorrect i'm afraid.

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Any chance I could get a pointer on where I went wrong with the first one?

#### Jhevon

MHF Helper
Any chance I could get a pointer on where I went wrong with the first one?
better yet, i gave you the solution

do you see where you went wrong? you forgot to put back the sec(5x) when you differentiated. and the sec(5x)tan(5x) is not the part that should be raised to the power

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#### [email protected]

The image didn't load the first time I viewed your response, I thereby withdraw my previous comment, and submit a new one in it's place:
Thanks a lot

I have another question, what software do you use to type in all the math formulas? It seems to work pretty well

#### Jhevon

MHF Helper
The image didn't load the first time I viewed your response, I thereby withdraw my previous comment, and submit a new one in it's place:
Thanks a lot
it was my bad, i forgot to post the image, when i went back and looked at the post i noticed it wasn't there, so i uploaded it then, sorry

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#### [email protected]

Related rates calculus question.

I'm having trouble getting the following word question:

A police officer in a patrol car is approaching an intersection at 25m/s. When he is 210 m from the intersection, a truck crosses the intersection travelling at right angles to the police car's path at a rate of 25m/s. If the oficer focusses his spotlight on the truck, how fast is the light beam turning 3 seconds later assuming that both vehicles continue at their same rates.

I would appreciate if someone could take a quick look at this question, maybe just give me a hint of what kind of approach to take, I have a feeling I can figure it out once I have a start.

The following two I thought I had figured out, but then I checked my answer with a graphing calculator and it didn't make any sense.

Find dy/dx of 1-2cos^2(x) = 4sin(x)cos(x)+y^2
I simplified this one to sin^2(x) = sin(4x)+y^2
and then tried to differentiate from there but it didn't work out, and I tried to differentiate the original as well. Was the original simplification incorrect?

Find dy/dx of (1/(1+cos(x)))
I tried this out and had something that looked good, but then I checked with a calculator and the derivative didn't mirror the slope of the equation at all.

Thanks for looking. Hopefully I can get all this stuff figured out soon and answer some questions for other people.

#### Jhevon

MHF Helper
I'm having trouble getting the following word question:

A police officer in a patrol car is approaching an intersection at 25m/s. When he is 210 m from the intersection, a truck crosses the intersection travelling at right angles to the police car's path at a rate of 25m/s. If the oficer focusses his spotlight on the truck, how fast is the light beam turning 3 seconds later assuming that both vehicles continue at their same rates.

I would appreciate if someone could take a quick look at this question, maybe just give me a hint of what kind of approach to take, I have a feeling I can figure it out once I have a start.
you asked for a hint, so that's what i'll give. see the diagram below, this is a related rates problem.

Find dy/dx of 1-2cos^2(x) = 4sin(x)cos(x)+y^2
I simplified this one to sin^2(x) = sin(4x)+y^2
and then tried to differentiate from there but it didn't work out, and I tried to differentiate the original as well. Was the original simplification incorrect?
yes, the simplification was incorrect.

1 - 2cos^2(x) = 4sin(x)cos(x) + y^2 ..........since cos(2x) = 2cos^2(x) - 1 = -(1 - 2cos^2(x)) and sin(2x) = 2sin(x)cos(x)
=> -cos(2x) = 2sin(2x) + y^2
=> y^2 = -cos(2x) - 2sin(2x)
try differentiating now

Find dy/dx of (1/(1+cos(x)))
y = 1/(1 + cos(x)) = (1 + cos(x))^-1
By the chain rule
=> dy/dx = -(1 + cos(x))^-2 * (-sin(x))
=> dy/dx = sin(x)(1 + cos(x))^-2

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I couldn't ask for better help, thank you!

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I'm working on that word problem currently. I would appreciate if someone could take a quick look at my solutions to the following problems. I have a feeling I messed something up, because I'm good at messing things up.

dy/dx of y=[csc^2(x)]-[3x^2][cot^2(x)]
dy/dx=2(cscx)d/dx(cscx) - 6x(2(cotx))d/dx(cotx)
dy/dx=2(cscx)(-cscxcotx) - 12x(cotx)(-csc^2(x))

dy/dx of y = csc((x^2-2)^3)
dy/dx= [-csc(x^2-2)^3][cot(x^2-2)^3][d/dx(x^2-2)^3]
dy/dx= [-csc(x^2-2)^3][cot(x^2-2)^3][3(x^2-2)^2][d/dx(x^2)]
dy/dx= [-csc(x^2-2)^3][cot(x^2-2)^3][3(x^2-2)^2][2x]