# I don't know what to do with these problems...

• December 8th 2008, 06:07 PM
NJL72413
I don't know what to do with these problems...
Hi, this is my first time using these forums, my brain is working at like half speed today for some reason... That and I can't remember what this is supposed to mean (Thinking)

1) not even going to bother with this one... don't have the time to type it out right now.

2) Find y[prime] for y^5 - x^5 = x^2 - y at the point (1,1)
A. 7/6

3) Find the equation of the tangent line to x = sin(t): y = 3 + t at t = pi/4. Place your answer in explicit form.
A. y = (sqrt2)x + 2 +pi/4

And I've found an answer close to the correct one for this question but I'm willing to bet that I've done it wrong (I did it without using calculus...), and I should be able to get the exact answer given.

4) One plane is flying south at 300 mph and is 5 miles north of a point. A second plane is flying east at 400 mph and is 12 miles west of the same point. How fast (to the nearest mph) are the places approaching each other?
A. 485 mph (the answer I got was 475.4 mph)

5) tan(x + y^2) = 1.612271 expresses the relationship between the number of thousands of miles (x) I've jogged since I was born and my weight (y) in tons. If I'm currently increasing my total mileage by 6 miles per year, what is my current weight and instantaneous weight loss in points/year? To date I have jogged a thousand miles in my life.
A. x = 1/8 tons, or 250 lbs

Thanks
• December 8th 2008, 06:16 PM
11rdc11
Quote:

Originally Posted by NJL72413
Hi, this is my first time using these forums, my brain is working at like half speed today for some reason... That and I can't remember what this is supposed to mean (Thinking)

1) not even going to bother with this one... don't have the time to type it out right now.

2) Find y[prime] for y^5 - x^5 = x^2 - y at the point (1,1)
A. 7/6

3) Find the equation of the tangent line to x = sin(t): y = 3 + t at t = pi/4. Place your answer in explicit form.
A. y = (sqrt2)x + 2 +pi/4

And I've found an answer close to the correct one for this question but I'm willing to bet that I've done it wrong (I did it without using calculus...), and I should be able to get the exact answer given.

4) One plane is flying south at 300 mph and is 5 miles north of a point. A second plane is flying east at 400 mph and is 12 miles west of the same point. How fast (to the nearest mph) are the places approaching each other?
A. 485 mph (the answer I got was 475.4 mph)

5) tan(x + y^2) = 1.612271 expresses the relationship between the number of thousands of miles (x) I've jogged since I was born and my weight (y) in tons. If I'm currently increasing my total mileage by 6 miles per year, what is my current weight and instantaneous weight loss in points/year? To date I have jogged a thousand miles in my life.
A. x = 1/8 tons, or 250 lbs

Thanks

$5y^4dy - 5x^4dx = 2xdx -dy$

$5y^4dy +dy = 2xdx + 5x^4dx$

$dy(5y^4 +1) = dx(2x + 5x^4)$

$\frac{dy}{dx} = \frac{2x + 5x^4}{5y^4 +1}$

now plug in your point (1,1)
• December 9th 2008, 09:48 AM
NJL72413
Thanks, I understand that one now, but I still can't make sense of the others =\

3) Find the equation of the tangent line to x = sin(t); y = 3 + t at t = pi/4. Place your answer in explicit form.
A. y = (sqrt2)x + 2 +pi/4

The equation of the tangent line is the derivative right? If that is true than I can take the derivative in parametric form, but what I get when I do that doesn't look anything like the answer. I'm guessing that I'm not even supposed to do it in parametric form but that's the only thing I can think of to do with it.

In parametric form I come out with...

y[prime]/x[prime] = 1/cos(t) and that looks nothing like the answer... what am I doing wrong, and am I doing anything right?
• December 9th 2008, 11:40 AM
Opalg
Quote:

Originally Posted by NJL72413
Thanks, I understand that one now, but I still can't make sense of the others =\

3) Find the equation of the tangent line to x = sin(t); y = 3 + t at t = pi/4. Place your answer in explicit form.
A. y = (sqrt2)x + 2 +pi/4

The equation of the tangent line is the derivative right? No. The derivative is the slope of the tangent line. If that is true than I can take the derivative in parametric form, but what I get when I do that doesn't look anything like the answer. I'm guessing that I'm not even supposed to do it in parametric form but that's the only thing I can think of to do with it.

In parametric form I come out with...

y[prime]/x[prime] = 1/cos(t) and that looks nothing like the answer... what am I doing wrong, and am I doing anything right?

So the slope of the tangent line at the point (sin(t),3+t) is 1/cos(t). Now you are told to put t=π/4 and then find the equation of the line with that slope going through that point. I'm sure you know how to do that.