Recent content by ugeous
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U
Finding the rate of change of an angle in triangle.
Great! Thank you Dick! :-)- ugeous
- Post #3
- Forum: Calculus and Beyond Homework Help
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U
Finding the rate of change of an angle in triangle.
Hello again! The base of an isosceles triangle is 20 cm and the altitude is increasing at the rate of 1 cm/min. At what rate is the base angle increasing when the area is 100 cm^2? I think I have found the solution, but want to have someone else check it over b/c I am not 100% sure...- ugeous
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- Angle Change Rate Rate of change Triangle
- Replies: 2
- Forum: Calculus and Beyond Homework Help
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U
Find max/min values of f(x)=cos(x)-sin(x)
Finally got it! Thank You!- ugeous
- Post #17
- Forum: Calculus and Beyond Homework Help
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U
Find max/min values of f(x)=cos(x)-sin(x)
Right!- ugeous
- Post #16
- Forum: Calculus and Beyond Homework Help
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U
Find max/min values of f(x)=cos(x)-sin(x)
-sqrt2/2, sqrt2/2 i believe.- ugeous
- Post #14
- Forum: Calculus and Beyond Homework Help
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U
Find max/min values of f(x)=cos(x)-sin(x)
This question is driving me crazy... I did find critical point, but I can't seem to find the second one. I know that usually you have more than one critical point when x can be more then one value, and still satisfy the equation. In this case, I had sin(x)=-cos(x). By dividing both sides by...- ugeous
- Post #12
- Forum: Calculus and Beyond Homework Help
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U
Find max/min values of f(x)=cos(x)-sin(x)
OK, I see. I used Dick's method (seems to be the fastest), and got the correct answer for maximum value(check with calculator), but how would I find the minimum value here? Equation that gave me max value was tan (x) = -1. Do I just need to use a different method for min value?- ugeous
- Post #10
- Forum: Calculus and Beyond Homework Help
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U
Find max/min values of f(x)=cos(x)-sin(x)
Find max/min values of f(x)=cos(x)-sin(x) in the interval of -pi (smaller or equals)x(smaller or equals)pi I took the derivative and ended up with f`(x)=-sin(x)-cos(x). By setting it equals to zero, i get sin(x)=-cos(x). This is where I'm stuck. Don't know where to go from here. thx!- ugeous
- Thread
- Replies: 16
- Forum: Calculus and Beyond Homework Help
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U
Trying to understand the question: Find dy/dx
Thanks man!- ugeous
- Post #5
- Forum: Calculus and Beyond Homework Help
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Trying to understand the question: Find dy/dx
Oh, right! sec(cube root x) * tan(cube root x) * 1/3(x^-2/3) Is this correct?- ugeous
- Post #3
- Forum: Calculus and Beyond Homework Help
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U
Trying to understand the question: Find dy/dx
Hello! Question asks: Find dy/dx. Do not simplify. y = sec(cube root x) My answer is dy/dx = sec(cube root x) * tan(cube root x) Am I correct? thanks!- ugeous
- Thread
- Replies: 4
- Forum: Calculus and Beyond Homework Help
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U
Evaluate each of the following limits. Show all reasoning.
just wanted to confirm with you guys (finding this topic a little difficult) by converting tan^2x i get lim 2sin^2x x->0 cos^2x(x^2) limit x->0 of sin^2x/x^2 gives 1 correct? then I am simply left with 2/cos^2x which gives me final answer 2. Am I correct? Thanks for help/comments.- ugeous
- Post #7
- Forum: Calculus and Beyond Homework Help
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U
Evaluate each of the following limits. Show all reasoning.
Ooooh... I have thought of a completely different thing here. I understand it now. thanks guys!- ugeous
- Post #6
- Forum: Calculus and Beyond Homework Help
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U
Evaluate each of the following limits. Show all reasoning.
Remind me please, do you rewrite tan as sin/cos before taking derivative or after? I think it is before, but not 100% sure. Thanks- ugeous
- Post #3
- Forum: Calculus and Beyond Homework Help
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U
Evaluate each of the following limits. Show all reasoning.
lim (2 tan^2 x)/x^2 x->0 I am just uncertain about how to do this question. Do I find a derivative or something else? Just need some guidance. Thanx.- ugeous
- Thread
- Limits
- Replies: 7
- Forum: Calculus and Beyond Homework Help