Magic car

Calculus Level 5

A toy car is placed in the x y xy plane at the coordinates ( 0 , 6 ) (0,6) . It faces the point ( 3 , 10 ) (3,10) , and is programmed to travel in a straight line in the direction of that point. If the height of the surface it travels through is modeled by the function f ( x , y ) = y sin ⁡ ( π x 2 ) + 2 x cos ⁡ ( π y 2 ) f(x,y)=y\sin(\frac{\pi x}{2})+2x\cos(\frac{\pi y}{2}) , and the car travels at at a constant speed of 10 10 units/second through the x y xy plane, calculate the slope (directional derivative) of the surface the car is at after 2 2 seconds.

If this slope can be expressed as S = a b ( c π + d ) S=\dfrac {a}{b}(c\pi+d) , where a a , b b , c c , and d d are integers with a a and b b being coprime positive integers and c c , a prime, enter a + b + c + d a+b+c+d .


The answer is 17.

Charley Shi by Charley Shi , 19, New Zealand

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1 solution

Chew-Seong Cheong
Apr 11, 2020

The directional vector of the toy car is v → = [ 3 − 0 10 − 6 ] = [ 3 4 ] \overrightarrow v = \begin{bmatrix} 3-0 \\ 10-6 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \end{bmatrix} and the unit vector is u → = [ 3 / 5 4 / 5 ] \overrightarrow u = \begin{bmatrix} 3/5 \\ 4/5 \end{bmatrix} .

Since the toy car moves at a constant speed of 10 10 units/second, this means that it moves 3 5 × 10 = 6 \frac 35 \times 10 = 6 units in the x x direction and 4 5 × 10 = 8 \frac 45 \times 10 = 8 units in the y y direction. Therefore after two seconds, the coordinates of the toy car is ( 0 + 2 × 6 , 6 + 2 × 8 ) = ( 12 , 22 ) (0+2\times 6, 6 + 2\times 8) = (12, 22) ; and the directional derivative is given by:

∇ u → f ( 12 , 22 ) = [ 3 5 4 5 ] ⋅ [ ∂ f ( x , y ) ∂ x ∂ f ( x , y ) ∂ y ] ( 12 , 22 ) = [ 3 5 4 5 ] ⋅ [ π y 2 cos ⁡ π x 2 + 2 cos ⁡ π y 2 sin ⁡ π x 2 − π x sin ⁡ π x 2 ] ( 12 , 22 ) = [ 3 5 4 5 ] ⋅ [ 11 π cos ⁡ ( 6 π ) + 2 cos ⁡ ( 11 π ) sin ⁡ ( 6 π ) − π x sin ⁡ ( 11 π ) ] = [ 3 5 4 5 ] ⋅ [ 11 π − 2 0 ] = 3 5 ( 11 π − 2 ) \begin{aligned} \nabla_{\overrightarrow u} f(12, 22) & = \begin{bmatrix} \frac 35 \\ \frac 45 \end{bmatrix} \cdot \begin{bmatrix} \frac {\partial f(x,y)}{\partial x} \\ \frac {\partial f(x,y)}{\partial y} \end{bmatrix}_{(12, 22)} \\ & = \begin{bmatrix} \frac 35 \\ \frac 45 \end{bmatrix} \cdot \begin{bmatrix} \frac {\pi y}2 \cos \frac {\pi x}2 + 2\cos \frac {\pi y}2 \\ \sin \frac {\pi x}2 - \pi x \sin \frac {\pi x}2 \end{bmatrix}_{(12, 22)} \\ & = \begin{bmatrix} \frac 35 \\ \frac 45 \end{bmatrix} \cdot \begin{bmatrix} 11\pi \cos (6\pi) + 2\cos (11\pi) \\ \sin (6\pi) - \pi x \sin (11\pi) \end{bmatrix} \\ & = \begin{bmatrix} \frac 35 \\ \frac 45 \end{bmatrix} \cdot \begin{bmatrix} 11\pi - 2 \\ 0 \end{bmatrix} \\ & = \frac 35 (11\pi - 2) \end{aligned}

Therefore a + b + c + d = 3 + 5 + 11 − 2 = 17 a+b+c+d = 3+5+11-2 = \boxed{17} .

2 Helpful 2 Interesting 4 Brilliant 1 Confused

@Charley Feng , we have to mention that a a and b b are positive coprime integers, because a = − 3 a=-3 and b = − 5 b=-5 also give the solution, then a + b + c + d = 1 a+b+c+d = 1 . Also note that 1 5 ( 33 π − 6 ) , 3 10 ( 22 π − 4 ) , 3 20 ( 44 π − 8 ) , ⋯ \dfrac 1{5}(33\pi -6), \dfrac 3{10}(22\pi - 4), \dfrac 3{20}(44\pi - 8), \cdots are also acceptable solutions. There are infinitely many. Therefore, we have to say that c c is a prime to fix the solution.

Chew-Seong Cheong - 1 year, 2 months ago

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