- #1
Kaldanis
- 106
- 0
Hi again, my calculus exam is on monday so trying hard to understand the things I'm struggling with. I think I'm almost there with part a - well I can solve it through drawing a graph or using a graphing program - but I'd rather solve it without graphs if possible. Part b however is causing me a lot of problems! Thanks in advance for any help
My attempt
The shortest line will be a straight perpendicular line, meaning the two gradients must multiply to give -1. I already know the first m=½, so the for the second line m=-2. Using m=-2, x=2 and y=-1 in the line equation, I get y=-2x+3. So now I have the equations of the two lines that are perpendicular to each other... but how do I work out the point at which they intersect? When graphing it I'm able to see visually straight away and then work out the distance, but I want to be able to do it without using a graphing program. Is there a way to work out the intersection point mathmatically?
My attempt
Problems like this are giving me the most problems. For the function to be continuous it just has to be defined everywhere, or basically it has to be able to be drawn without taking the pen off the paper. So the function is indeed continuous? To work out if it's differentiable, well I can differentiate both of the equations but I don't think it's as easy as that? Then to find extrema I set the derivative equal to zero and see when this happens?
(a) Let Q be the point in the x-y-plane with coordinates (2,−1) and let L be the line given by the equation y = ½x+3.
Find the coordinates of the point P on the line L such that the distance between P and Q is minimal. Calculate this distance.
Find the coordinates of the point P on the line L such that the distance between P and Q is minimal. Calculate this distance.
My attempt
The shortest line will be a straight perpendicular line, meaning the two gradients must multiply to give -1. I already know the first m=½, so the for the second line m=-2. Using m=-2, x=2 and y=-1 in the line equation, I get y=-2x+3. So now I have the equations of the two lines that are perpendicular to each other... but how do I work out the point at which they intersect? When graphing it I'm able to see visually straight away and then work out the distance, but I want to be able to do it without using a graphing program. Is there a way to work out the intersection point mathmatically?
(b) Let g be the function on the interval [−3, 5] defined by [tex]g(x) = \begin{cases} x^{3}+4 & \text{ if } -3 \le x \le -1 \\ -x^{2}-5x-1 & \text{ if } -1 < x \le 5 \end{cases}[/tex]Determine whether g is continuous and whether g is differentiable on [−3, 5], determine the absolute maxima and minima of g on the interval [−3, 5], and determine the corresponding maximal and minimal values of g.
My attempt
Problems like this are giving me the most problems. For the function to be continuous it just has to be defined everywhere, or basically it has to be able to be drawn without taking the pen off the paper. So the function is indeed continuous? To work out if it's differentiable, well I can differentiate both of the equations but I don't think it's as easy as that? Then to find extrema I set the derivative equal to zero and see when this happens?