Use tangent to find area of triangle

[Plutonium88]

Homework Statement

the tangent of the graph 1/x^2 at P(2,1/4) forms a triangle with the x and y axis. Find area of triangle.

Homework Equations

The Attempt at a Solution

so f'(x)=-2/x^3

so the slope of the tangent at point 2, is f'(2)=-2/8 = -1/4=mt

now i use the equation of the line to determine x + y intercepts

y=mtx+b plug in point P(2,1/4)
1/4=-1/4(2) + b
b=3/4 (y intercept_)

y=-1/4x+3/4
let y=0
0=-1/4x+3/4
x=3

therefore AreaΔ=bxh/2

where Base = |b| = 3/4
where height = |x| = 3

A=(3/4)(3)/2
A=9/8 units^2I believe this is the correct solution, to me everything makes sense, but I'm just kind of nervous about the area.. 9/8 units i mean i know its possible but i dunno... I'm worried I'm messing my numbers up or something.

***i found a mistake where my slope was 1/4 and it was supposed to be -1/4

 

[Ibix]

Have you tried plotting your tangent line and the curve on graph paper? Does the tangent line look plausible, and does the old counting-the-squares-inside-the-triangle method give you something like 9/8? I think you are right, but checks you can do yourself always reassure.

You switch between x and t for the horizontal coordinate, and m and mt for the gradient. You seem to come out of it with the right answer (now - I spotted your sign error but when I clicked "quote" it had gone, which caused a moment of doubt of my sanity), so I guess these are typos. Proofreading needed when you're going to hand in for real...

 

[eumyang]

Looks fine to me. Just a couple of things to nitpick:

so f'(x)=-2/x^3

so the slope of the tangent at point 2, is f'(-2)=-2/8 = -1/4=mt

Change the "-2" to "2" and "mt" to "m".

now i use the equation of the line to determine x + y intercepts

y=mt+b plug in point P(2,1/4)

Change "mt" to "mx".

 

[Plutonium88]

Looks fine to me. Just a couple of things to nitpick:

Change the "-2" to "2" and "mt" to "m".


Change "mt" to "mx".

okay thanks, i already did the recalculation when x=2, not -2, i just forgot to change the function but i edited it back. and i made it y=Mtx+b is that okay?

 

[Plutonium88]

Have you tried plotting your tangent line and the curve on graph paper? Does the tangent line look plausible, and does the old counting-the-squares-inside-the-triangle method give you something like 9/8? I think you are right, but checks you can do yourself always reassure.

You switch between x and t for the horizontal coordinate, and m and mt for the gradient. You seem to come out of it with the right answer (now - I spotted your sign error but when I clicked "quote" it had gone, which caused a moment of doubt of my sanity), so I guess these are typos. Proofreading needed when you're going to hand in for real...

yea sorry about that, and yea i did graph it, it just was looking a little tight on if the tangent could reach point 3 but it's definately possible. Thanks for the help and the ideas for checking