Find the equation of the tangent in the given problem

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In summary, the textbook solution is indicated below: to find the slope of the tangent line to the y-axis at (x, y) = (b, 0), use the equation y = -8x+12. When x = 0, y = 4. When x = b, y = 2.
  • #1
chwala
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Homework Statement
See attached
Relevant Equations
Differentiation
1666782546246.png
The textbook solution is indicated below;

1666782657824.png


My approach on this question is as follows;

##\dfrac{dy}{dx} = -\dfrac {8}{x^3}##

##\dfrac{dy}{dx} (x=a) = -\dfrac {8}{a^3}##

The tangent equation is given by;

##y= -\dfrac {8}{a^3}x+\dfrac{12+a^2}{a^2}##

when ##x=0##,

##y=\dfrac{12+a^2}{a^2}##

and when ##y=0##,

##\dfrac {8}{a^3} x = \dfrac{12+a^2}{a^2}##

##⇒\dfrac {8}{a^2}=y##

##⇒\dfrac {8}{a^2}=\dfrac{4}{a^2} +1##

##8=4+a^2##

##4=a^2##

##a=\sqrt{4}=2##

on using; ##y=\dfrac{12+a^2}{a^2}## and substituting ##a=2## yields

##y=\dfrac{12+2^2}{2^2}=\dfrac{16}{4}=4##

##b=4##.

I would appreciate any other method or insight. Cheers guys!
 
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  • #2
What happened to ##a=-2## at ##y=0##?

I think your method is fine, no need to make things more complicated.
 
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  • #3
chwala said:
Homework Statement:: See attached
Relevant Equations:: Differentiation

View attachment 316098The textbook solution is indicated below;

View attachment 316099

My approach on this question is as follows;

##\dfrac{dy}{dx} = -\dfrac {8}{x^3}##

##\dfrac{dy}{dx} (x=a) = -\dfrac {8}{a^3}##

The tangent equation is given by;

##y= -\dfrac {8}{a^3}x+\dfrac{12+a^2}{a^2}##

when ##x=0##,

##y=\dfrac{12+a^2}{a^2}##

and when ##y=0##,

##\dfrac {8}{a^3} x = \dfrac{12+a^2}{a^2}##

##⇒\dfrac {8}{a^2}=y##

##⇒\dfrac {8}{a^2}=\dfrac{4}{a^2} +1##

##8=4+a^2##

##4=a^2##

##a=\sqrt{4}=2##

on using; ##y=\dfrac{12+a^2}{a^2}## and substituting ##a=2## yields

##y=\dfrac{12+2^2}{2^2}=\dfrac{16}{4}=4##

##b=4##.

I would appreciate any other method or insight. Cheers guys!
My only thought is that, once you get the slope, the line that passes through the point (0, b) is ##(y - b) = m(x - 0 )##, and the line through (b, 0) is ##(y - 0) = m(x - b)##. You can solve those simultaneously. However I don't think that approach gets you anything that much simpler.

-Dan
 
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fresh_42 said:
What happened to ##a=-2## at ##y=0##?

I think your method is fine, no need to make things more complicated.
You are right; ##a=±2##, We shall treat ##a=-2## as unsuitable as it will not realize the required co-ordinates at the ##x## axes, that is ##(b,0)## and ##y## axes, that is ##(0,b)##.
 
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  • #5
The conditions on the tangent line intersecting the axes require that its equation be [tex]
f(a) + f'(a)(x - a) = \frac{12}{a^2} + 1 - \frac{8}{a^3}x = y = b - x.[/tex] Comparing powers of [itex]x[/itex] gives [tex]
\begin{split}
\frac{12}{a^2} + 1 &= b \\
\frac{8}{a^3} &= 1.\end{split}[/tex] These are easily solved to find [itex]a = 2[/itex] and [itex]b = 4[/itex].
 
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FAQ: Find the equation of the tangent in the given problem

1. What does it mean to find the equation of the tangent?

Finding the equation of the tangent means determining the mathematical formula that describes the line that touches a curve at a specific point, known as the point of tangency. This equation allows us to calculate the slope of the curve at that point, which is the rate of change of the curve at that specific point.

2. How do you find the equation of the tangent?

To find the equation of the tangent, we need to first find the slope of the curve at the point of tangency. This can be done by taking the derivative of the function that represents the curve. Then, we use the point-slope form of a line to write the equation, using the coordinates of the point of tangency and the slope we calculated.

3. Why is finding the equation of the tangent important?

Knowing the equation of the tangent allows us to understand the behavior of a curve at a specific point. It also helps us determine the rate of change of the curve at that point, which can be useful in many applications, such as physics, engineering, and economics.

4. Can the equation of the tangent change at different points on the curve?

Yes, the equation of the tangent can change at different points on the curve because the slope of the curve can vary at different points. This means that the rate of change of the curve is not constant and can change depending on the location on the curve.

5. Are there any special cases when finding the equation of the tangent?

Yes, there are some special cases when finding the equation of the tangent. For example, if the curve is a straight line, the equation of the tangent will be the same as the equation of the curve itself. Additionally, if the curve has a vertical tangent, the equation of the tangent will not exist because the slope is undefined at that point.

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