Forum for the GRE subject test in mathematics.

CoCoA
 Posts: 42
 Joined: Wed Sep 03, 2008 5:39 pm
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by CoCoA » Sun Sep 07, 2008 3:15 pm
If f(x)=\int_0^x{x^2sin(t^2)dt}, then what is f'(x)?

CoCoA
 Posts: 42
 Joined: Wed Sep 03, 2008 5:39 pm
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by CoCoA » Sun Sep 07, 2008 11:02 pm
I have not solved this. Here is a simpler one, but with a good idea.
Let a curve be defined parametrically by
x=\int_1^t{\frac{cos u}{u}du},
y=\int_1^t{\frac{sin u}{u}du}.
Find the arc length from the origin to the nearest point on the curve with a vertical tangent.

Nameless
 Posts: 128
 Joined: Sun Aug 31, 2008 4:42 pm
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by Nameless » Mon Sep 08, 2008 7:41 am
If f(x)=\int_0^x{x^2sin(t^2)dt}, then what is f'(x)?
Can we use the formula
f(x)=\int_a(x)^b(x){g(t)dt}>f'(x)=g[b(x)]b'(x)g[a(x)]a'(x)

Nameless
 Posts: 128
 Joined: Sun Aug 31, 2008 4:42 pm
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by Nameless » Mon Sep 08, 2008 7:52 am
I have not solved this. Here is a simpler one, but with a good idea.
Let a curve be defined parametrically by
x=\int_1^t{\frac{cos u}{u}du},
y=\int_1^t{\frac{sin u}{u}du}.
Find the arc length from the origin to the nearest point on the curve with a vertical tangent.
the arc length l=int_0^a(sqrt[x'(t)^2+[y'(t)^2])dt
if x=x(t)
y=y(t) then y'(x)=dy/dx=y'(t)/x'(t)> from this equation solve for a
and calculate the integral

CoCoA
 Posts: 42
 Joined: Wed Sep 03, 2008 5:39 pm
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by CoCoA » Mon Sep 08, 2008 12:17 pm
Can we use the formula
f(x)=\int_a(x)^b(x){g(t)dt}>f'(x)=g[b(x)]b'(x)g[a(x)]a'(x)
Not when there is a function of x in the integrand.

CoCoA
 Posts: 42
 Joined: Wed Sep 03, 2008 5:39 pm
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by CoCoA » Tue Sep 09, 2008 1:30 am
the arc length l=int_0^a(sqrt[x'(t)^2+[y'(t)^2])dt
if x=x(t)
y=y(t) then y'(x)=dy/dx=y'(t)/x'(t)> from this equation solve for a
and calculate the integral
Very close, but one detail needs to be changed.

CoCoA
 Posts: 42
 Joined: Wed Sep 03, 2008 5:39 pm
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by CoCoA » Thu Sep 25, 2008 12:11 am
Still must finish. Go ahead & do it (you will have to do it twice) then tell me how you resolve the integral with only t in it