Multivariable Chain Rule - Multivariable Calculus

1. Let $F(x,y,z) = 0$ define $z$ implicitly as a function of $x$ and $y$. If $F_x = 6$, $F_y = -4$, and $F_z = 8$, what is $\frac{\partial z}{\partial x}$?
   Answer: $-\frac{3}{4}$ When you see an equation like $F(x,y,z) = 0$ that defines $z$ implicitly as a function of $x$ and $y$, you need to use implicit differentiation to find partial derivatives like $\frac{\partial z}{\partial x}$. The key insight is applying the chain rule. When you differentiate $F(x,y,z) = 0$ with respect to $x$ while treating $y$ as constant, you get: $\frac{\partial F}{\partial x} + \frac{\partial F}{\partial z} \cdot \frac{\partial z}{\partial x} = 0$ This comes from recognizing that $z$ depends on $x$, so differentiating $F$ with respect to $z$ requires multiplying by $\frac{\partial z}{\partial x}$. Solving for $\frac{\partial z}{\partial x}$: $\frac{\partial z}{\partial x} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}} = -\frac{F_x}{F_z}$ Substituting the given values: $\frac{\partial z}{\partial x} = -\frac{6}{8} = -\frac{3}{4}$ Answer D is correct. Answer A ($\frac{4}{3}$) uses the wrong ratio and misses the negative sign. Answer B ($\frac{3}{4}$) has the correct magnitude but forgets the crucial negative sign that comes from moving $F_x$ to the other side of the equation. Answer C ($-\frac{4}{3}$) incorrectly uses $\frac{F_z}{F_x}$ instead of $\frac{F_x}{F_z}$ but does include the negative sign. Remember the formula: $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$ and $\frac{\partial z}{\partial y} = -\frac{F_y}{F_z}$. The negative sign is essential and comes directly from implicit differentiation.
2. Let $z$ be a differentiable function of $x$ and $y$ that satisfies the equation $z^3 + zx + y^2 = 12$. If $x(t) = 2t^2$ and $y(t) = 3t$, find the value of $\frac{dz}{dt}$ at the moment when $t=1$.
   Answer: $-\frac{22}{5}$ First, find the values of $x$, $y$, and $z$ at $t=1$. At $t=1$, we have $x(1)=2(1)^2=2$ and $y(1)=3(1)=3$. Substituting these into the governing equation gives $z^3 + z(2) + (3)^2 = 12$, which simplifies to $z^3 + 2z - 3 = 0$. By inspection, $z=1$ is a solution. To find $\frac{dz}{dt}$, we use the chain rule: $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$. We need the partial derivatives of $z$. Differentiating the equation $z^3 + zx + y^2 = 12$ implicitly: With respect to $x$: $3z^2\frac{\partial z}{\partial x} + (z + x\frac{\partial z}{\partial x}) + 0 = 0 \implies (3z^2+x)\frac{\partial z}{\partial x} = -z \implies \frac{\partial z}{\partial x} = \frac{-z}{3z^2+x}$. With respect to $y$: $3z^2\frac{\partial z}{\partial y} + x\frac{\partial z}{\partial y} + 2y = 0 \implies (3z^2+x)\frac{\partial z}{\partial y} = -2y \implies \frac{\partial z}{\partial y} = \frac{-2y}{3z^2+x}$. At the point $(x,y,z)=(2,3,1)$, we have $\frac{\partial z}{\partial x} = \frac{-1}{3(1)^2+2} = -\frac{1}{5}$ and $\frac{\partial z}{\partial y} = \frac{-2(3)}{3(1)^2+2} = -\frac{6}{5}$. Next, we find the derivatives of the path functions: $\frac{dx}{dt} = 4t$ and $\frac{dy}{dt} = 3$. At $t=1$, $\frac{dx}{dt}=4$ and $\frac{dy}{dt}=3$. Finally, substitute all values into the chain rule: $\frac{dz}{dt} = (-\frac{1}{5})(4) + (-\frac{6}{5})(3) = -\frac{4}{5} - \frac{18}{5} = -\frac{22}{5}$.
3. Let $w = f(x,y,z)$ where $x = u + v$, $y = u - v$, and $z = uv$. If $\frac{\partial w}{\partial u} = 3$ and $\frac{\partial w}{\partial v} = -2$ at a certain point, what is $\frac{\partial w}{\partial x}$ at that point?
   Answer: $\frac{1}{2}$ This is a classic chain rule problem involving implicit functions. When you have a function $w = f(x,y,z)$ where $x$, $y$, and $z$ are themselves functions of other variables, you need to use the chain rule to find partial derivatives with respect to the new variables. To find $\frac{\partial w}{\partial x}$, you need to express it in terms of the given information about $\frac{\partial w}{\partial u}$ and $\frac{\partial w}{\partial v}$. Since $x = u + v$, you can solve for $u$ and $v$ in terms of $x$ and $y$: from $x = u + v$ and $y = u - v$, you get $u = \frac{x+y}{2}$ and $v = \frac{x-y}{2}$. Using the chain rule: $\frac{\partial w}{\partial x} = \frac{\partial w}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial w}{\partial v}\frac{\partial v}{\partial x}$ From the expressions above, $\frac{\partial u}{\partial x} = \frac{1}{2}$ and $\frac{\partial v}{\partial x} = \frac{1}{2}$. Substituting: $\frac{\partial w}{\partial x} = 3 \cdot \frac{1}{2} + (-2) \cdot \frac{1}{2} = \frac{3}{2} - 1 = \frac{1}{2}$ Option A ($1$) likely comes from incorrectly assuming $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial x} = 1$. Option B ($\frac{5}{2}$) results from adding the given partial derivatives directly. Option D ($\frac{3}{2}$) comes from using only the first term of the chain rule. Remember: when variables are related through transformations, always write out the full chain rule expansion and carefully compute each partial derivative in the transformation.
4. If $z = \ln(x^2 + y^2)$ where $x = e^t \cos s$ and $y = e^t \sin s$, what is $\frac{\partial z}{\partial s}$?
   Answer: $0$ First, note that $x^2 + y^2 = e^{2t}\cos^2 s + e^{2t}\sin^2 s = e^{2t}(\cos^2 s + \sin^2 s) = e^{2t}$. Therefore $z = \ln(e^{2t}) = 2t$. Since $z$ depends only on $t$ and not on $s$, we have $\frac{\partial z}{\partial s} = 0$. The other choices result from incorrectly applying the chain rule without first simplifying the expression.
5. Let $z = f(x, y)$, where $x = r \cos\theta$ and $y = r \sin\theta$. Given that $\frac{\partial z}{\partial r} = \frac{1}{r}x$ and $\frac{\partial z}{\partial \theta} = -y$, find the expression for $\frac{\partial z}{\partial x}$ in terms of $x$ and $y$.
   Answer: $1$ When you encounter a problem involving coordinate transformations like polar to Cartesian, you need to use the chain rule to connect partial derivatives in different coordinate systems. To find $\frac{\partial z}{\partial x}$, use the chain rule: $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial z}{\partial \theta} \frac{\partial \theta}{\partial x}$ First, find the needed partial derivatives. From $r = \sqrt{x^2 + y^2}$, we get $\frac{\partial r}{\partial x} = \frac{x}{r}$. From $\theta = \arctan(y/x)$, we get $\frac{\partial \theta}{\partial x} = -\frac{y}{x^2 + y^2}$. Substituting the given information: $\frac{\partial z}{\partial x} = \frac{1}{r}x \cdot \frac{x}{r} + (-y) \cdot \left(-\frac{y}{x^2 + y^2}\right) = \frac{x^2}{r^2} + \frac{y^2}{x^2 + y^2}$ Since $r^2 = x^2 + y^2$, this becomes: $\frac{\partial z}{\partial x} = \frac{x^2}{x^2 + y^2} + \frac{y^2}{x^2 + y^2} = \frac{x^2 + y^2}{x^2 + y^2} = 1$ The answer is D. Option A gives $\frac{2x}{x^2+y^2}$, which would result from incorrectly doubling the first term. Option B suggests $x^2$, ignoring the coordinate transformation entirely. Option C gives $\frac{x}{x^2+y^2}$, which results from forgetting to include the second term from the chain rule. Remember: when transforming between coordinate systems, always apply the full chain rule and carefully track all terms—they often combine to give surprisingly simple results.
6. Let $z = f(x,y)$ where $x = g(u,v)$ and $y = h(u,v)$. If $f_x = 3$, $f_y = -2$, $g_u = 4$, $g_v = -1$, $h_u = 2$, and $h_v = 5$, what is $\frac{\partial z}{\partial v}$?
   Answer: $-13$ Using the multivariable chain rule: $\frac{\partial z}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial v} = f_x \cdot g_v + f_y \cdot h_v = 3(-1) + (-2)(5) = -3 + (-10) = -13$. Choice B incorrectly omits the second term. Choice C uses incorrect signs. Choice D uses the absolute value.
7. Let $w = f(x^2 - 4y^2)$, where $f$ is any differentiable function of a single variable. Which of the following partial differential equations must $w$ satisfy for any choice of $f$?
   Answer: $4y \frac{\partial w}{\partial x} + x \frac{\partial w}{\partial y} = 0$ Let $u = x^2 - 4y^2$, so $w = f(u)$. We use the chain rule to find the partial derivatives of $w$ with respect to $x$ and $y$. Note that $f$ is a function of a single variable $u$, so its derivative is $f'(u)$. $\frac{\partial w}{\partial x} = \frac{df}{du} \frac{\partial u}{\partial x} = f'(u) \cdot (2x)$. $\frac{\partial w}{\partial y} = \frac{df}{du} \frac{\partial u}{\partial y} = f'(u) \cdot (-8y)$. To find a relationship that holds for any $f$, we must eliminate the $f'(u)$ term. We can do this by solving for $f'(u)$ in both equations: From the first equation, $f'(u) = \frac{1}{2x} \frac{\partial w}{\partial x}$ (for $x\neq 0$). From the second equation, $f'(u) = \frac{1}{-8y} \frac{\partial w}{\partial y}$ (for $y\neq 0$). Setting the expressions for $f'(u)$ equal gives: $\frac{1}{2x} \frac{\partial w}{\partial x} = \frac{1}{-8y} \frac{\partial w}{\partial y}$. Multiplying both sides by $-16xy$ to clear denominators gives: $-8y \frac{\partial w}{\partial x} = 2x \frac{\partial w}{\partial y}$. Rearranging the terms to one side gives the partial differential equation: $8y \frac{\partial w}{\partial x} + 2x \frac{\partial w}{\partial y} = 0$, which simplifies to $4y \frac{\partial w}{\partial x} + x \frac{\partial w}{\partial y} = 0$.
8. If $u = f(x-y, y-z, z-x)$ where $f$ is differentiable, which expression represents $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z}$?
   Answer: $0$ Let $w_1 = x-y$, $w_2 = y-z$, $w_3 = z-x$. Then $\frac{\partial u}{\partial x} = f_1 \cdot 1 + f_2 \cdot 0 + f_3 \cdot (-1) = f_1 - f_3$, $\frac{\partial u}{\partial y} = f_1 \cdot (-1) + f_2 \cdot 1 + f_3 \cdot 0 = -f_1 + f_2$, $\frac{\partial u}{\partial z} = f_1 \cdot 0 + f_2 \cdot (-1) + f_3 \cdot 1 = -f_2 + f_3$. Adding: $(f_1 - f_3) + (-f_1 + f_2) + (-f_2 + f_3) = 0$. The other choices result from incorrectly applying the chain rule.
9. Let $z = f(x, y)$ be a differentiable function. A particle moves in the $xy$-plane along a path $\vec{r}(t) = \langle x(t), y(t) \rangle$. It is observed that the value of $z$ remains constant along the particle's path. Which of the following statements must be true?
   Answer: The particle's path follows a level curve of the function $f$. The rate of change of $z$ along the path is given by the chain rule: $\frac{dz}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$. This can be written as the dot product of the gradient of $f$, $\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \rangle$, and the velocity vector of the particle, $\vec{v}(t) = \vec{r}'(t) = \langle \frac{dx}{dt}, \frac{dy}{dt} \rangle$. So, $\frac{dz}{dt} = \nabla f \cdot \vec{v}(t)$. We are given that $z$ remains constant along the path, which means $\frac{dz}{dt} = 0$. Therefore, $\nabla f \cdot \vec{v}(t) = 0$. This implies that the gradient vector $\nabla f$ is orthogonal (perpendicular) to the velocity vector $\vec{v}(t)$ at every point on the path. The velocity vector is tangent to the path. Since the gradient vector is always normal to the level curves of a function, and the path's tangent vector is always orthogonal to the gradient, the path must be tangent to the level curves at every point. This means the particle's path follows a level curve of $f$.
10. Let $P(V, T)$ be the pressure of a gas, depending on volume $V$ and temperature $T$. The volume and temperature are changing with time $t$ according to $V(t) = 10 + 0.5t^2$ and $T(t) = 300 + 2t$. At time $t=10$, it is measured that $V=60$, $T=320$, $\frac{\partial P}{\partial V} = -2$, and $\frac{\partial P}{\partial T} = 5$. However, the volume measurements are calibrated in liters (L) and the temperature in Kelvin (K), while the time is in seconds (s). The pressure is measured in kilopascals (kPa). What is the rate of change of pressure with respect to time, $\frac{dP}{dt}$, at $t=10$ seconds?
    Answer: $-10$ kPa/s The rate of change of pressure with respect to time is given by the chain rule: $\frac{dP}{dt} = \frac{\partial P}{\partial V}\frac{dV}{dt} + \frac{\partial P}{\partial T}\frac{dT}{dt}$. We are given the values of the partial derivatives at the time of interest: $\frac{\partial P}{\partial V} = -2$ kPa/L and $\frac{\partial P}{\partial T} = 5$ kPa/K. We need to find the rates of change of volume and temperature with respect to time by differentiating their respective functions. $V(t) = 10 + 0.5t^2 \implies \frac{dV}{dt} = t$. $T(t) = 300 + 2t \implies \frac{dT}{dt} = 2$. Now, we evaluate these rates at $t=10$ s. $\frac{dV}{dt}|_{t=10} = 10$ L/s. $\frac{dT}{dt}|_{t=10} = 2$ K/s. The given values of $V=60$ and $T=320$ at $t=10$ confirm we are at the correct point, as $V(10) = 10+0.5(100)=60$ and $T(10)=300+2(10)=320$. We can now substitute all the values into the chain rule expression: $\frac{dP}{dt} = (-2 \text{ kPa/L})(10 \text{ L/s}) + (5 \text{ kPa/K})(2 \text{ K/s}) = -20 \text{ kPa/s} + 10 \text{ kPa/s} = -10 \text{ kPa/s}$.
11. Using the information provided, what is the value of $\frac{\partial w}{\partial r}$ at the point $(r, \theta) = (2, \pi/6)$?
    Answer: $7(\sqrt{3}-1)$ The chain rule for $\frac{\partial w}{\partial r}$ requires us to first find $\frac{\partial w}{\partial x}$ and $\frac{\partial w}{\partial y}$. $\frac{\partial w}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} = (6)(3) + (-2)(2) = 18 - 4 = 14$. $\frac{\partial w}{\partial y} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial y} = (6)(-1) + (-2)(4) = -6 - 8 = -14$. Next, we find the partial derivatives of $x$ and $y$ with respect to $r$: $x = r\cos\theta \implies \frac{\partial x}{\partial r} = \cos\theta$. $y = r\sin\theta \implies \frac{\partial y}{\partial r} = \sin\theta$. At $(r, \theta) = (2, \pi/6)$, $\frac{\partial x}{\partial r} = \cos(\pi/6) = \frac{\sqrt{3}}{2}$ and $\frac{\partial y}{\partial r} = \sin(\pi/6) = \frac{1}{2}$. Now, we can apply the final chain rule: $\frac{\partial w}{\partial r} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial r}$. Substituting the values we found: $\frac{\partial w}{\partial r} = (14)(\frac{\sqrt{3}}{2}) + (-14)(\frac{1}{2}) = 7\sqrt{3} - 7 = 7(\sqrt{3}-1)$. Distractor B comes from a sign error when computing $\frac{\partial w}{\partial y}$. Distractor C comes from a cross-term error when computing $\frac{\partial w}{\partial x}$. Distractor D is the value of $\frac{\partial w}{\partial \theta}$.
12. Let $w = g(u, v)$, where $u = x+y$ and $v = \frac{y}{x}$. It is known that at the point $(x,y)=(1,1)$, we have $u=2, v=1$, and the partial derivatives are $\frac{\partial g}{\partial u} = 4$ and $\frac{\partial g}{\partial v} = -2$. What is the value of $\frac{\partial w}{\partial x}$ at $(x,y)=(1,1)$?
    Answer: $6$ When you encounter a problem involving a composite function where the inner variables are functions of other variables, you need to use the chain rule for multivariable functions. Here, $w = g(u,v)$ where both $u$ and $v$ depend on $x$ and $y$. To find $\frac{\partial w}{\partial x}$, apply the multivariable chain rule: $\frac{\partial w}{\partial x} = \frac{\partial g}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial g}{\partial v} \cdot \frac{\partial v}{\partial x}$ First, calculate the partial derivatives of the inner functions: - $u = x + y$, so $\frac{\partial u}{\partial x} = 1$ - $v = \frac{y}{x}$, so $\frac{\partial v}{\partial x} = -\frac{y}{x^2}$ At the point $(x,y) = (1,1)$, we have $\frac{\partial v}{\partial x} = -\frac{1}{1^2} = -1$. Now substitute the given values: $\frac{\partial g}{\partial u} = 4$ and $\frac{\partial g}{\partial v} = -2$: $\frac{\partial w}{\partial x} = 4 \cdot 1 + (-2) \cdot (-1) = 4 + 2 = 6$ Choice A ($2$) would result from forgetting the second term of the chain rule. Choice C ($4$) comes from using only the first term $\frac{\partial g}{\partial u} \cdot \frac{\partial u}{\partial x}$. Choice D ($8$) might result from sign errors or incorrect application of the quotient rule. Study tip: For chain rule problems, always write out the full formula first, then systematically find each partial derivative. Double-check your derivative of $v = \frac{y}{x}$ with respect to $x$ — treat $y$ as constant and use the power rule: $v = yx^{-1}$.
13. Which of the following expressions is equivalent to $\frac{\partial^2 w}{\partial x \partial y}$?
    Answer: $\frac{\partial^2 f}{\partial u^2} - \frac{\partial^2 f}{\partial v^2}$ When you encounter a change of variables problem in multivariable calculus, you need to systematically apply the chain rule to transform partial derivatives from one coordinate system to another. To find $\frac{\partial^2 w}{\partial x \partial y}$, start by finding the first-order partial derivatives. Using the chain rule with $u = x + y$ and $v = x - y$: $\frac{\partial w}{\partial x} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} = \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}$ Now differentiate this with respect to $y$: $\frac{\partial^2 w}{\partial x \partial y} = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}\right)$ Applying the chain rule again: $= \frac{\partial^2 f}{\partial u^2} \cdot \frac{\partial u}{\partial y} + \frac{\partial^2 f}{\partial u \partial v} \cdot \frac{\partial v}{\partial y} + \frac{\partial^2 f}{\partial v \partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial^2 f}{\partial v^2} \cdot \frac{\partial v}{\partial y}$ Since $\frac{\partial u}{\partial y} = 1$ and $\frac{\partial v}{\partial y} = -1$, and mixed partials are equal ($\frac{\partial^2 f}{\partial u \partial v} = \frac{\partial^2 f}{\partial v \partial u}$): $= \frac{\partial^2 f}{\partial u^2} - \frac{\partial^2 f}{\partial u \partial v} + \frac{\partial^2 f}{\partial v \partial u} - \frac{\partial^2 f}{\partial v^2} = \frac{\partial^2 f}{\partial u^2} - \frac{\partial^2 f}{\partial v^2}$ This confirms answer D. A incorrectly assumes mixed partials aren't equal. B omits the negative signs from the chain rule. C includes an extra mixed partial term that should cancel out. Study tip: Always track the signs carefully when applying the chain rule, and remember that mixed partial derivatives are equal for smooth functions.
14. The equation $x \ln y + y \ln z + z \ln x = 0$ implicitly defines $z$ as a function of $x$ and $y$ near the point $(1, 1, 1)$. Find the value of $\frac{\partial z}{\partial x}$ at this point.
    Answer: $-1$ Let $F(x, y, z) = x \ln y + y \ln z + z \ln x = 0$. The formula for implicit differentiation is $\frac{\partial z}{\partial x} = -\frac{\partial F / \partial x}{\partial F / \partial z}$. First, we compute the partial derivatives of $F$. $\frac{\partial F}{\partial x} = \frac{\partial}{\partial x}(x \ln y + y \ln z + z \ln x) = \ln y + 0 + \frac{z}{x}$. $\frac{\partial F}{\partial z} = \frac{\partial}{\partial z}(x \ln y + y \ln z + z \ln x) = 0 + \frac{y}{z} + \ln x$. Now, we evaluate these partial derivatives at the point $(1, 1, 1)$. $\frac{\partial F}{\partial x}|_{(1,1,1)} = \ln(1) + \frac{1}{1} = 0 + 1 = 1$. $\frac{\partial F}{\partial z}|_{(1,1,1)} = \frac{1}{1} + \ln(1) = 1 + 0 = 1$. Finally, we use the formula: $\frac{\partial z}{\partial x} = -\frac{1}{1} = -1$. A common error is to forget the negative sign in the formula, leading to the answer 1.
15. The temperature on a metal plate is given by $T(x, y) = 100 - x^2 - 3y^2$. An ant walks along the elliptical path $\vec{r}(t) = \langle 2\cos(t), \sin(t) \rangle$ for $t \ge 0$. What is the rate of change of the temperature experienced by the ant at the first moment it crosses the line $y = x$?
    Answer: $\frac{4}{5}$ The rate of change of temperature is $\frac{dT}{dt} = \frac{\partial T}{\partial x}\frac{dx}{dt} + \frac{\partial T}{\partial y}\frac{dy}{dt}$. First, find the derivatives: $\frac{\partial T}{\partial x} = -2x$, $\frac{\partial T}{\partial y} = -6y$, $\frac{dx}{dt} = -2\sin(t)$, $\frac{dy}{dt} = \cos(t)$. Substituting the path equations $x=2\cos(t)$ and $y=\sin(t)$ into the chain rule gives: $\frac{dT}{dt} = (-2(2\cos t))(-2\sin t) + (-6(\sin t))(\cos t) = 8\sin t\cos t - 6\sin t\cos t = 2\sin t\cos t = \sin(2t)$. The ant crosses the line $y=x$ when $\sin(t) = 2\cos(t)$, which means $\tan(t)=2$. The first time this occurs for $t \ge 0$ is in the first quadrant. Let this time be $t_0$. We can find the values of $\sin(t_0)$ and $\cos(t_0)$ from a right triangle with opposite side 2, adjacent side 1, and hypotenuse $\sqrt{5}$. So, $\sin(t_0) = \frac{2}{\sqrt{5}}$ and $\cos(t_0) = \frac{1}{\sqrt{5}}$. The rate of change at $t_0$ is $\frac{dT}{dt}|_{t=t_0} = 2\sin(t_0)\cos(t_0) = 2(\frac{2}{\sqrt{5}})(\frac{1}{\sqrt{5}}) = \frac{4}{5}$.

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