Equation of tangent line w/ sqrt.

In summary, the homework statement is to find the equation of the tangent line to y=sqrt(25-x^2) at the point (4,3). The Attempt at a Solution suggests that you use the chain rule to find f'(x), which is the slope at any x. The equation of the tangent line is f'(4)=4(25-4^2)^-1/2.
  • #1
Calc.Hatr10
6
0

Homework Statement


Find the equation of the tangent line to y=sqrt(25-x^2) at the point (4,3)


Homework Equations





The Attempt at a Solution


(25-x^2)^1/2
1/2(25-x^2)^-1/2
(2x)1/2(25-x^2)^-1/2
x(25-x^2)^-1/2

I have no idea what to do or where to go after this...
Please Help!
 
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  • #2
so is that chain rule differntition you tried? very hard to follow...

if you have a slope & point, you should be able to come up with a an equation of a line no worries
 
  • #3
yes I tried the chain rule...and I can't seem to find the slope
 
  • #4
you were given f(x), using chain rule you have found f'(x) which is the slope at any x, try substituting in the x valuue of the point you are given
 
  • #5
the point is (4,3)

f'(x)= x(25-x^2)^-1/2
f'(4)= 4(25-4^2)^-1/2
f'(4)= 4(25+16)^-1/2
f'(4)= 4(41)^-1/2
 
Last edited:
  • #6
lost a negative
 
  • #7
Im checking over my work thus far and i cannot see where I lost a neg.
at (25-4^2)
-4^2=+16


Please explain
 
  • #8
it think it should be (25-(4^2))
 
  • #9
[tex](-4)^2 = (-1*4)^2 = (-1)^2*(4)^2 = 16[/tex]

[tex]-(4^2) = (-1)*4^2 = -16[/tex]
 
  • #10
so..
f'(4)=4(25-(4^2))^-1/2
f'(4)=4(25-16)^-1/2
f'(4)=4(9)^-1/2
f'(4)=4(1/3)
f'(4)=4/3

?
 
  • #11
You made a slight error with the derivative.

[tex]y=(25-x^2)^{1/2}[/tex]

[tex]\frac{dy}{dx}=\frac{1}{2}(25-x^2)^{-1/2}*(-2x)[/tex]

[tex]\frac{dy}{dx}=-x(25-x^2)^{-1/2}[/tex]

Can you take it from here?
 
  • #12
put the x value into you equation for y', to get the slope, then just find the line with that slope and the point you initially had
 
  • #13
Find tangent line to y= sqrt(25-x^2) at the point (4,3)

f'(x)= (25-x^2)^1/2
f'(x)= 1/2(25-x^2)^-1/2
f'(x)= (-2x)1/2(25-x^2)^-1/2
f'(x)= -x(25-x^2)^-1/2

f'(4)=-4(25-4^2)^-1/2
f'(4)=-4(25-(4^2))^-1/2
f'(4)=-4(25-16)^-1/2
f'(4)=-4(9)^-1/2
f'(4)=-4(1/3)
f'(4)=-4/3

3=-4/3(4)+b
3=-16/3+b
0=(-16/3)/3+b
0=-16/9+b
-b=-16/9
b=16/9

y= -4/3x+16/9

please tell me I am done with this problem
 
  • #14
Calc.Hatr10 said:
Find tangent line to y= sqrt(25-x^2) at the point (4,3)

3=-16/3+b
0=(-16/3)/3+b

The second line above is wrong. What you should do is add 16/3 to both sides so you get b by itself. After that, the equation of the tangent line will be correct.
You can check if you have the correct tangent line on a graphing calculator, or if you don't have one, look for one online.
 
  • #15
Calc.Hatr10 said:
3=-4/3(4)+b
3=-16/3+b
0=(-16/3)/3+b
0=-16/9+b
-b=-16/9
b=16/9

please tell me I am done with this problem

Just follow Bohrok's advice and you are done with this problem. However, judging by these few lines of working, it is evident that you are VERY innefficient in algebra manipulation and equation solving. Before you started calc you should've known how to solve equations properly, since you'll be doing so a lot throughout the course and it will be impossible for you to pass if you don't already know these prerequisites inside out.

Sorry, but it's just how it is. You need to go back and learn algebra and equation solving (even more-so).

For e.g.

3=-16/3+b
0=(-16/3)/3+b

You've attempted to divide both sides by 3. Instead, you only divide one term on the right-hand side and ignore the b (probably because it made things look easier for you). Also, when you divided the left-hand side by 3, you resulted in the number 0? 3/3=1, 10/10=1, a/a=1 if a[itex]\neq[/itex]0.

Good luck!
 

1. What is the equation of the tangent line for a function with a square root?

The equation of the tangent line for a function with a square root is given by y = mx + b, where m is the slope of the tangent line and b is the y-intercept. To find the slope, you can use the derivative of the function. Then, you can use the point-slope form of a line to find the equation of the tangent line.

2. How do I find the slope of the tangent line for a square root function?

The slope of the tangent line for a square root function can be found by taking the derivative of the function and evaluating it at the point of tangency. This will give you the slope of the tangent line at that point.

3. Do I need to use the chain rule when finding the equation of the tangent line for a square root function?

Yes, you will need to use the chain rule when finding the equation of the tangent line for a square root function. This is because the square root function is a composition of two functions - the square root function and the inside function. So, when you take the derivative of the square root function, you will need to use the chain rule to find the derivative of the inside function.

4. Can I use the point-slope form of a line to find the equation of the tangent line for a square root function?

Yes, you can use the point-slope form of a line to find the equation of the tangent line for a square root function. This form is y - y1 = m(x - x1), where m is the slope of the tangent line and (x1, y1) is the point of tangency. You can find the slope using the derivative of the function and then plug in the coordinates of the point of tangency to find the equation of the tangent line.

5. Are there any special cases when finding the equation of the tangent line for a square root function?

Yes, there are a few special cases to consider when finding the equation of the tangent line for a square root function. If the function has a vertical tangent line at a certain point, the slope of the tangent line will be undefined and you will need to use the vertical line equation x = a to find the equation of the tangent line. Additionally, if the point of tangency is the x-intercept of the function, the equation of the tangent line will be y = 0.

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