MHB Jack's question at Yahoo Answers regarding finding gradients

  • Thread starter Thread starter MarkFL
  • Start date Start date
AI Thread Summary
To find the gradients of tangents to the curve f(x)=ln(x+1) that form a 45-degree angle with the tangent at x=1, the derivative f'(x) is calculated as f'(x)=1/(x+1). Evaluating this at x=1 gives f'(1)=1/2. The possible gradients m are determined by setting the angle difference to π/4, leading to the equation |tan^(-1)(m) - tan^(-1)(1/2)| = π/4. Solving this results in the gradients m = -1/3 and m = 3, confirming that the tangents are perpendicular.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

LOGARITHMIC FUNCTION, MATHS QUESTION PLEASEE HELPP ME IM BEGGIN YOU!?

Consider the curve f(x)=ln(x+1). find the gradients of the possible tangents to f(x) which makes of 45 degrees with the tangent of f(x) at the point where x=1

Please explain your answer thankyou

I have posted a link there to this topic so the OP can see my work.
 
Mathematics news on Phys.org
Hello Jack,

First, we want to find the gradient of the line tangent to the given logarithmic curve where $x=1$. To do so, we must differentiate the curve with respect to $x$:

$$f'(x)=\frac{1}{x+1}$$

Now, we evaluate this for $x=1$:

$$f'(1)=\frac{1}{2}$$

Then, to find the possible gradients $m$ that make an angle of 45° with the tangent line at its point of tangency, we may equate the magnitude of the difference in the angles of inclination to $$\frac{\pi}{4}$$.

$$\left|\tan^{-1}(m)-\tan^{-1}\left(\frac{1}{2} \right) \right|=\frac{\pi}{4}$$

$$\tan^{-1}(m)-\tan^{-1}\left(\frac{1}{2} \right)=\pm\frac{\pi}{4}$$

Taking the tangent of both sides, we find:

$$\tan\left(\tan^{-1}(m)-\tan^{-1}\left(\frac{1}{2} \right) \right)=\tan\left(\pm\frac{\pi}{4} \right)$$

Using the angle-difference identity for tangent on the left, and simplifying the right, we obtain:

$$\frac{m-\frac{1}{2}}{1+\frac{m}{2}}=\pm1$$

Now we may solve for $m$. Multiply through by $$2\left(1+\frac{m}{2} \right)$$

$$2m-1=\pm(2+m)$$

Square both sides, then arrange as the difference of squares:

$$(2m-1)^2-(2+m)^2=0$$

Apply the difference of squares formula:

$$(2m-1+2+m)(2m-1-2-m)=0$$

Combine like terms:

$$(3m+1)(m-3)=0$$

Hence the possible gradients are:

$$m=-\frac{1}{3},3$$

As we should expect, the product of the two gradients is -1 as the two lines would be perpendicular to one another.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top