# Use tangent to find area of triangle

## 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.

## 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^2

I beleive 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

Last edited:

Related Calculus and Beyond Homework Help News on Phys.org
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...

Last edited:
eumyang
Homework Helper
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".

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?

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 definitly possible. Thanks for the help and the ideas for checking