If, $H(x,y,z)=2x^2-yz+xz^2,x=2\sin t,y=t^2-t-1, z=\frac{3}{e}$ Find at t=0, $\frac{dH}{dt}$.

Question

I'm just checking my working if it's correct. The answer given in my textbook is 24, but I think there's a problem with that.

Here's the question. If, $$H(x,y,z)=2x^2-yz+xz^2,x=2\sin t,y=t^2-t-1, z=\frac{3}{e}$$

Find at t=0, $\frac{dH}{dt}$.

Here's my working, $$\frac{dx}{dt}=2\cos t,\frac{dy}{dt}=2t-1,\frac{dx}{dt}=0$$

So,

$$\frac{dH}{dt}=\frac{\partial H}{\partial x}\frac{dx}{dt}+\frac{\partial H}{\partial y}\frac{dy}{dt}+\frac{\partial H}{\partial z}\frac{dz}{dt}$$

$$\frac{dH}{dt}=\frac{\partial H}{\partial x}\frac{dx}{dt}+\frac{\partial H}{\partial y}\frac{dy}{dt}+0$$

$$\frac{dH}{dt}=(4x+z^2)(2\cos t)+(-z)(2t-1)$$

$$t=0, \cos t=1, x=0, y=-1, z=\frac{3}{e}$$

$$\frac{dH}{dt}=(4(0)+\frac{3^2}{e^2})(2\cos 0)+(-\frac{3}{e})(2(0)-1)$$

$$\frac{dH}{dt}=\frac{18}{e^2}+\frac{3}{e}$$

Is my answer correct? Or am I missing out something which prevented me from getting the answer in my textbook (24)?

Answer 1

Your solution is correct, so there’s either a typo in the original problem or the answer key is wrong. This wouldn’t be the first time that has happened. Even if you eliminate the factor of $1/e$ from from $z$, you will end up with $21$, not $24$.