Really hard math question

[waterloggedfrog]

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This is my last question! 😀 I ran into two problems like this on my first pearson practice test, it's a math problem where you have a triangle with a right angle inside of a circle and then it asks for dy/dx of one of the sides of the triangle. No lengths are given, just x,y,etc. The answer explanation they give is just no help at all. I can't figure it out! Any help would be great, I just need to be pointed in the right direction. I tried googling "triangle inside circle" to no avail

 

[SYPHARMACY]

what is the question?

 

[diastole]

I saw that when I was taking it and said to myself that I was immediately skipping a question like that on the exam if I saw one. While it is fine to learn how to do it now, if it is going to be a question that takes extra mental power on test day, skip it and go back to it if you have time.

 

[Pharmpills]

This is my last question! 😀 I ran into two problems like this on my first pearson practice test, it's a math problem where you have a triangle with a right angle inside of a circle and then it asks for dy/dx of one of the sides of the triangle. No lengths are given, just x,y,etc. The answer explanation they give is just no help at all. I can't figure it out! Any help would be great, I just need to be pointed in the right direction. I tried googling "triangle inside circle" to no avail

They have to give you some value like the radius of the circle or length of a side, arc, etc bc otherwise it would be impossible to solve and get an actual value but you could solve it and get an equation where you simply plug in the values when you get it.

 

[waterloggedfrog]

I was going to skip it but then the practice test had two questions like that so I figured I should learn how to do it. The question says that R is constant and x and y is changing and it asks to find dy/dx when x=r/2

 

[uncbiograd]

i've been working on this problem for the last 30ish minutes 😡
I have the first half down, you use x^2+(1/2y)^2=r^2
then take the derivative so you have 2x+y/2(dy/dx)=0
long story short you solve for dy/dx

the part i don't understand is how you get from this dy/dx answer "-4x/y" to any of the answers given. it says that you know when x=r/2 the ration x/y is something, but i don't know how they got that...

edit: ok so once you figure out dy/dx = -4x/y

then assume that x=1 and r=2 since you are given x=r/2
once again use x^2 + (1/2y)^2 = r ^2 with those values and you solve to find that y = 2sqrt(3)

finally insert x=1 and y=2sqrt(3) into -4x/y and you get the answer for dy/dx

note to self: don't obsess over a problem like this on the pcat! haha this would have taken the majority of the quant section for me. ugh.

 

[waterloggedfrog]

Wow, it makes sense now! Thanks a lot. Now maybe if it's on the pcat though it won't take us 30 minutes and we can get a sure point!

 

[Aquahelix]

Just remember the answer from the practive test !

 

[stkyrice]

honestly, those questions are designed to deliberately be time wasters. you're much better off answering the easier questions and racking up points that way than wasting 4-5 minutes on one question that you probably will get wrong.

i got my 99 composite (back in aug 07) because my math score was 25 points higher than any of my other scores, and i know i didn't answer all the questions, especially the ones that involved a lot of calculus, like chain rule, differential equations, and some integral calculus.