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my calculus final is tomorrow and after going over my calculus review there are a few problems I could really use some help with.
1) Use local linear approximation at x=8 with the function f(x) = x^(1/3) to approximate 7.9^(1/3)
I know that local linear approximation looks something like this:
L(x) = f'(a)*(x-a)+f(a)
I don't know what I'm supposed to plug in for x and a.
2) A ball is thrown upward in the air from the edge of a building 96 feet tall with an initial velocity of 80 ft/sec. How fast is the ball moving when it hits the ground?
I'm pretty stumped on this one, how do I figure out what the acceleration is to plug into a velocity function?
3) Find the right Riemann sum with n=3 for the function f(x) = x^3 + 1 for x= -1 to x=2.
Is this really just the right endpoints for each integer from -1 to 2 plugged into the function?
my calculus final is tomorrow and after going over my calculus review there are a few problems I could really use some help with.
1) Use local linear approximation at x=8 with the function f(x) = x^(1/3) to approximate 7.9^(1/3)
I know that local linear approximation looks something like this:
L(x) = f'(a)*(x-a)+f(a)
I don't know what I'm supposed to plug in for x and a.
2) A ball is thrown upward in the air from the edge of a building 96 feet tall with an initial velocity of 80 ft/sec. How fast is the ball moving when it hits the ground?
I'm pretty stumped on this one, how do I figure out what the acceleration is to plug into a velocity function?
As far as 2) is concerned, it's the acceleration due to gravity. 9.81mm^2 I believe. Hopefully that's all you need to do that problem.
Do you know about taylor series? If so 1. Is just the first two terms of a taylor series around x=7.9, evaluated at a=8. just plug those into your formula and go
Usually those kinds of problems use a flat 32 in ft/s^2 and 9.8 m/s^2.
The first one is basically using the tangent line to approximate the function. So plug in the value you want to estimate the function at into the tangent line equation.
Three is just as easy as you think it is.
enlightenedbum on
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Usually those kinds of problems use a flat 32 in ft/s^2 and 9.8 m/s^2.
The first one is basically using the tangent line to approximate the function. So plug in the value you want to estimate the function at into the tangent line equation.
Three is just as easy as you think it is.
Thanks, that helped me a lot with those problems!
There is one more that I thought I could do but apparently am having a hard time with.
Evaluate the integral } x cos(x^2) dx
Can anyone explain how to do that? I'm very good at using u substitution but I don't see how it would work for this one...
Usually those kinds of problems use a flat 32 in ft/s^2 and 9.8 m/s^2.
The first one is basically using the tangent line to approximate the function. So plug in the value you want to estimate the function at into the tangent line equation.
Three is just as easy as you think it is.
Thanks, that helped me a lot with those problems!
There is one more that I thought I could do but apparently am having a hard time with.
Evaluate the integral } x cos(x^2) dx
Can anyone explain how to do that? I'm very good at using u substitution but I don't see how it would work for this one...
Let u = x^2 , du = 2x dx
Then x cos(x^2) dx = 1/2 cos u du and I think you can take it from there
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As far as 2) is concerned, it's the acceleration due to gravity. 9.81mm^2 I believe. Hopefully that's all you need to do that problem.
OP is using ft/sec and not m/s, so the acceleration is 32.2 feet/sec^2
The first one is basically using the tangent line to approximate the function. So plug in the value you want to estimate the function at into the tangent line equation.
Three is just as easy as you think it is.
Thanks, that helped me a lot with those problems!
There is one more that I thought I could do but apparently am having a hard time with.
Evaluate the integral } x cos(x^2) dx
Can anyone explain how to do that? I'm very good at using u substitution but I don't see how it would work for this one...
PSN: rlinkmanl
Let u = x^2 , du = 2x dx
Then x cos(x^2) dx = 1/2 cos u du and I think you can take it from there