Recent content by tc903

Are the given families orthogonal trajectories of each other? $$ {x}^{2}+{y}^{2} = ax $$ $$ {x}^{2}+{y}^{2} = by $$ I first started by finding them implicitly. $$ \frac{2x+a}{2y} = y' $$ $$ \frac{2x+b}{2y} = y' $$ Then the problem wanted me to sketch my answer. The Tschirnhausen, I...

$$ F = k{R}^{4} $$ The flux F is volume of blood per unit time. This is proportional to the 4th power of the radius R of the blood vessel. All I am given is 3% increase in radius will affect blood flow how. I am to find whether is decreases or increase blood flow and by what percent. $$...

Thanks MarkFl, you were extremely helpful and our answers agree.

I had to leave. Did you mean $$ a{x}^{2}+bx+c = 0 $$ $$ {b}^{2} - 4ac = 0 $$ So that I may find repeated real solutions. $$ {x}^{2} + 4\left({(mx)}^{2}-12{m}^{2}x+3mx-12{m}^{2}+144{m}^{2}-36m+3mx-36m+9\right) = 36 $$ Skipping steps shown. $$...

The second problem. $$ y - {y}_{1} = m\left(x-{x}_{1}\right) $$ $$ y - 3 = m\left(x-12\right) $$ $$ y = m\left(x-12\right) + 3 $$Substituting the above,$$ {x}^{2} + 4{y}^{2} = 36 $$ $$ {x}^{2} +4\left(m\left(x-12\right)+3\right)^{2} = 36 $$ $$ {x}^{2} + 4\left(mx-12m+3\right)^{2} = 36 $$

Alright, here is my work. $$ f(x) = \arctan\left({\frac{\sqrt{1+x}}{\sqrt{1-x}}}\right) $$ $$ \d{f}{x} = \frac{1}{1+{\left(\frac{\sqrt{1+x}}{\sqrt{1-x}}\right)}^{2}}\left[\d{}{x}{\left(\frac{1+x}{1-x}\right)}^{1/2}\right] [/MATh] $$ \d{f}{x} =...

I know what I missed. Thanks MarkFL, I will get back on the second problem soon.

I found $$\d{f}{x}$$ for the first one both using quotient rule and product rule. I changed the latex so it showed my answer now.

Given$$ f(x) = \arctan\left({\frac{\sqrt{1+x}}{\sqrt{1-x}}}\right)$$ I differentiated and this was my answer. $$\d{y}{x} = \frac{1}{2\sqrt{1+x}\sqrt{1-x}{(1-x)}^{2}}$$ I used implicit differentiation on the elliptic curve $${x}^{2}+4{y}^{2} = 36$$ and it wants two horizontal tangents through...

My first question states y=a^3+cos^3 (x) (I couldn't quite figure out latex again.) The derivative using the chain rule I found to be y'=(a^3)(ln a)+3((cos (x))^2)(-sin(x)) = y' = (a^3)(ln a)-3((cos (x))^2)(sin (x)) The second question y=[x+(x+sin^2 (x))^7]^5 Derived using chain rule...

I would think the same, maybe I shouldn't have rounded it to -15.9 when I originally did that way. I have a computer that is telling me no when I plug in an answer like that.

I should have tried that. I just didnt think it was an option because it didnt say I can put it in their. The question usually asks if the limit exists find it. (type so and so for inifinity or for none exist) It had only chosen one of those. Thanks MarkFL!

I stated it did not exist, but that is wrong. I substituted zero and that would be make it undefined but that is wrong. It isn't - \infty . It is zero? Here is a table. x,y -1,-6 -2,-3 -3,-2 -4,-3/2 -5,-6/5 -6,-1 -7,-6/7 -8,-3/4 -9,-2/3 -10,-3/5

Thanks MarkFL. I was wondering if you can clear something up with the second question though with the absolute value.

H=16t-1.86t^2 is the height formula. The ball was thrown 16 m/s. I derived 16-3.72a. When will the rock hit the surface. It hits the surface 8.6 seconds. What velocity does the rock hit the surface? H=0.0344 then the velocity is 15.87 m/s? This is what I chose. [-4.-2) (-2,2) [2,4) (4,6) (6,8)