Common Tangent of Two Curves

For the sake of variety, let me suggest a purely algebraic solution, without reference to calculus.

Let's write the common tangent as $L(x)=mx+b$ . The discriminant of $x^2-L(x)=x^2-mx-b$ must be 0, meaning that $m^2+4b=0$ and $b=-\frac{m^2}{4}$ .

The discriminant of $x^3-L(x)=x^3-mx+\frac{m^2}{4}$ must be 0 as well, meaning that $4m^3-\frac{27m^4}{16}=0$ . Thus the nonzero value of $m$ is $m=\frac{64}{27}$ and the answer is $\boxed{91}$ .

The gradients of the two curves are given by

$\begin{cases} m_1 = \dfrac {dy}{dx} = 2x \\ m_2 = \dfrac {dy}{dx} = 3x^2 \end{cases}$

The tangents at $(a_1, a_1^2)$ and $(a_2, a_2^3)$ of curve 1 and curve 2 are

$\begin{cases} \dfrac {y-a_1^2}{x-a_1} = 2a_1 & \implies y = 2a_1x - a_1^2 \\ \dfrac {y-a_2^3}{x-a_2} = 3a_2^2 & \implies y = 3a_2^2x - 2a_2^3 \end{cases}$

For tangent 1 and tangent 2 to be same, we equate the gradients and the $y$ -intercept $\begin{cases} 2a_1 = 3a_2^2 & ...(1) \\ a_1^2 = 2a_2^3 & ...(2) \end{cases}$ . From $\dfrac {(2)}{(1)}: \dfrac {a_1}2 = \dfrac {2a_2}3$ $\implies a_2 = \frac 34 a_1$ . From $(1): 2a_1 = 3 \left(\frac 34 a_1\right)^2$ $\implies a_1 = \frac {32}{27}$ . Since gradient of the common tangent $m_1 = m_2 = 2a_1 = \frac {64}{27}$ , $p+q = 64+27 = \boxed{91}$ .

Let the equation of the tangent be $y=mx+c$

Let the points of contact on $y=x^2$ and $y=x^3$ be $(a,a^2)$ and $(b,b^3)$ respectively.

Now, $m=2a$ and $m=3b^2$ . (Slope of tangent = $f'(x)$ )

Also, the tangent passes through the above two points,

$a^2=ma+c$

$b^3=mb+c$

Subtracting the above two equations,

$a^2-b^3=m(a-b)$

Now, just plug in the values of $a$ and $b$ in terms of $m$ , and cancel out $m^{\frac {3}{2}}$ (thus cancelling out m=0), and you should get the value of $m$ as $\frac {64}{27}$

Suppose the line is tangent to $f(x)=x^3$ in $x_1$ and is tangent to $g(x)=x^2$ in $x_2$ . Then the slope of the tangent can be expressed in three ways: $f'(x_1)=g'(x_2)=\frac{g(x_2)-f(x_1)}{x_2-x_1}$ . So $3x_1^2=2x_2=\frac{x_2^2-x_1^3}{x_2-x_1}$ .

Substitute $x_2= \frac{3}{2}x_1^2$ to get

$3x_1^2=\frac{\frac{9}{4}x_1^4-x_1^3}{\frac{3}{2}x_1^2-x_1}$ . Since $x_1\neq 0$ we can simplify this to

$3 (6x_1^2-4x_1)=9x_1^2-4x_1$ and further to

$9x_1-8=0$

$x_1=\frac{8}{9}$

The slope then is $3x_1^2=64/27$ , and our answer is $64+27=91$

Let $f(x) = x^2$ and $g(x) = x^3$ and $a$ and $b$ be real numbers.

$f'(a) = 2a = g'(b) = 3b^2 \implies a = \dfrac{3}{2}b^2$ .

$A:(a,a^2) = (\dfrac{3}{2}b^2,\dfrac{9}{4}b^4)$ and $B:(b,b^3) \implies m_{AB} = \dfrac{b^2(9b - 4)}{2(3b - 2)} = 3b^2$ and $b \neq 0 \implies b = \dfrac{8}{9} \implies m_{AB} = \dfrac{64}{27} = \dfrac{p}{q} \implies p + q = \boxed{91}$ .

What does "(in sky blue)" mean?

Let the intersection of the tangent line and $y=x^3$ be at $(a,a^3)$ . Then the tangent line is in the form $y-a^3=3a^2(x-a)$ . This tangent line must intersect $y=x^2$ , and be tangent to it: $m(=3a^2)=2x$ . Solving these $3$ equations for $m$ , we get $m = \frac{64}{27}$ .

Let the points of tangency be "a" on x^3 and "b" on x^2. Then the slope of the tangent line for x^3 with tangency point "a" is f'(a)=3a^2. Similarly, the slope of the tangent line for x^2 with tangency point b is f'(b)=2b.

So we can write the linear tangent lines in the form of y=(3a^2)x+c and y=2bx+d, where c and d are constants for constant values of a and b. For (3a^2)x+c to be tangent to x^3 at a, it follows that the tangency point must be (a,a^3).

This implies (3a^2) a+c=a^3. Then 3a^3+c=a^3, so c=-2a^3. We can use the same logic for the tangent line 2bx+d. The tangency point will be (b,b^2), so 2b b+d=b^2, which will mean d=-b^2.

So our tangent lines are (3a^2)x-2a^3, and 2bx-b^2. In order for the tangent lines to be the same line, their slopes and constants must be equal. This means 2b=3a^2 and -b^2=-2a^3. So b^2=2a^3 and 2b=3a^2.

Consider dividing the second equation by the first. Then b^2/2b=2a^3/3a^2, which results in b/2=2a/3, which means b=4a/3. Plugging this into our second equation gives us 2*4a/3=3a^2, so 8a/3=3a^2, so 3a=8/3, so a=8/9.

Now we can plug in the value of "a" into 3a^2 (the slope of the tangent line), to find the slope is 3*64/81, giving us the slope to be 64/27=p/q. Since 64 and 27 are relatively prime, our answer is 64+27= 91 .

Tangent to curve y=f(x) at point 'a' is given by

y=f(a)+(x-a){(d/dx)(f(x)}@a.

Applying,

(1)Tangent to y=x^3 at x1 is

y=3(x1^2)x - 2(x1)^3

(2)Tangent to y=x^2 at x2 is

y=2(x2)x - (x2)^2

Equating,

3(x1)^2=2(x2)

and

2(x1)^3=(x2)^2,

we get x1=8/9 and x2=32/27.

The slope = 2(x2) = 3(x1^2) = 64/27.

Answer=91