find the equations of the tangent line and normal line?

Question

to the curve y=f(x) at the point having given value of x.

1. f(x)=(2x-3)^(5/2), x=2
2. f(x)=xsqrt(25-x^2), x=4
3. f(x) = (sqrt(5+2x))/x, x=2

Can I please have an explanation to what a normal line is and how to solve these step by step? Thanks

Anonymous · Accepted Answer

The normal line is a line that is perpendicular to the tangent line and passes through the same point on f(x) as the tangent line does.

I will solve the first one for you. The second and third problems are similar!

1)

Taking derivatives gives:

f'(x) = (2)(5/2)(2x - 3)^(5/2 - 1)
==> f'(x) = 5(2x - 3)^(3/2).

At x = 2, we see that the slope of the tangent line is:

f'(2) = 2[2(2) - 3]^(5/2) = 2(1)^(5/2) = 2.

(This gives the slope of the normal line to be -1/2)

Since f(x) passes through (2, 1), we see that the tangent line is:

y - 1 = 2(x - 2)
==> y = 2x - 3.

Then, the normal line is:

y - 1 = (-1/2)(x - 2)
==> y = (-1/2)x + 2.

I hope this helps!

fireFox · Answer

find the coordinates of the given point.  then find the value of f'(x) at the given point; this is the slope of the tangent line at that point.  the normal line is perpendicular to the tangent line.  its slope is the negative reciprocal of the slope of the tangent line, or -1/f'(x).  then with point and slope you can determine the equation of the line.

for example
f(x) = (2x - 3)^(5/2), x = 2
at x = 2
f(2) = (2*2 - 3)^(5/2)
f(2) = 1
thus the point is (2,1)

f'(x) = 5/2*(2x - 3)^(3/2)*2
at x = 2
f'(x) = 5/2*(2*2 - 3)^(3/2)*2
f'(2) = 5

tangent
(y - 1)/(x - 2) = 5
normal
(y - 1)/(x - 2) = -1/5

simplify the equations; the rest you can do yourself

find the equations of the tangent line and normal line?

2 Answers