Calculus III Advanced (Course) (13.5) (Homework)
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Section 13.5 Homework
From Calculus 10e by Larson and Edwards, p. 913. Exercises 12, 16, 20, 21.
Exercise 13.5.12 Projectile Motion
The parametric equations for the paths of two projectiles are given. At what rate is the distance between the two objects changing at the given value for \(t\)?
| \(x_{1}\) | \(=48 \sqrt{2}t, \:\:\:\: \) | \(y_{1} \) | \( = 48 \sqrt{2}t -16t^{2} \:\:\:\: \) | First Object |
| \(x_{2}\) | \(=48 \sqrt{3} t, \) | \(y_{2} \) | \( = 48t-16t^{2} \) | Second Object |
| \(t\) | \(= 1 \) |
Solution The change rate is the \(dt\) at time \(t=1\) set equal to the distance formula. Therefore,
| \(dt \) | \(= \sqrt{ (x_{1}(t)-x_{2}(t))^{2} + (y_{1}(t)-y_{2}(t))^{2}} \) |
| \(dt^{2} \) | \(= (x_{1}(t)-x_{2}(t))^{2} + (y_{1}(t)-y_{2}(t))^{2} \) |
Take the derivative on both sides with respect to \(t\) and use the chain rule to produce
- \( 2dd^{\prime} = 2(x_{1}(t)-x_{2}(t))(x_{1}^{\prime}(t)-x_{2}^{\prime}(t)) + 2(y_{1}(t)-y_{2}(t))(y_{1}^{\prime}(t)-y_{2}^{\prime}(t)) \)
- $$ d^{\prime} t = \frac{ (x_{1}(t)-x_{2}(t))(x_{1}^{\prime}(t)-x_{2}^{\prime}(t)) + (y_{1}(t)-y_{2}(t))(y_{1}^{\prime}(t)-y_{2}^{\prime}(t)) }{dt} $$
When \(t=1\)
| \(x_{1}(1)\) | \(= 48\sqrt{2}, \:\:\:\: \) | \(x_{1}^{\prime}(1)\) | \(= 48\sqrt{2} , \:\:\:\: \) | \(y_{1}(1)\) | \(= 48\sqrt{2}-16, \:\:\:\: \) | \(y_{1}^{\prime}(1)\) | \(= 48\sqrt{2}-32 \) |
| \(x_{2}(1)\) | \(= 48\sqrt{3}, \:\:\:\: \) | \(x_{2}^{\prime}(1)\) | \(= 48\sqrt{3}, \:\:\:\: \) | \(y_{2}(1)\) | \(= 32, \:\:\:\: \) | \(y_{2}^{\prime}(1)\) | \(= 16 \) |
- \(d(1) = 48 \sqrt{ (\sqrt{2} -\sqrt{3})^{2} + (\sqrt{2}-1)^{2} } \)
Substitute the data into \(d^{\prime}(t) \) produces
| \(d^{\prime}(1) \) | $$= \frac{ 48^{2} (\sqrt{2} -\sqrt{3})^{2} + 48^{2}(\sqrt{2}-1)^{2}}{48 \sqrt{ (\sqrt{2} -\sqrt{3})^{2} + (\sqrt{2}-1)^{2} }}$$ |
| \(= 48 \sqrt{ (\sqrt{2} -\sqrt{3})^{2} + (\sqrt{2}-1)^{2}} \) | |
| \(= 48 \sqrt{ 8- 2\sqrt{6} - 2\sqrt{2} } \) |
Exercise 13.5.16 Finding Partial Derivatives
Find \(\partial w/\partial s\) and \(\partial w/\partial t\) using the appropriate Chain Rule. Evaluate each partial derivative at the given values for \(s\) and \(t\).
| Function | Values |
|---|---|
| \( w=x^{2}-y^{2} \) | \(s=3, \: t = \frac{\pi}{4} \) |
| \( x=s \cos t, \: y = s \sin t \:\:\:\: \) |
Solution
| \(x(3,\pi/4) \) | \(= 3 \cos(\pi/4) = 3 / \sqrt{2} \) |
| \(y( 3, \pi/4 ) \) | \(= 3 \sin (\pi/4) = 3 / \sqrt{2} \) |
- $$ \frac{\partial w}{ \partial s} = \frac{\partial w}{ \partial x} \frac{\partial x}{ \partial s} + \frac{\partial w}{ \partial y}\frac{\partial y}{ \partial s} = 2x \cdot \cos t -2y \cdot \sin t $$
- $$ = 2 \cdot 3 / \sqrt{2} \cdot 3 / \sqrt{2} - 2 \cdot 3 / \sqrt{2} \cdot 3 / \sqrt{2} = 0 $$
- $$ \frac{\partial w}{ \partial t} = \frac{\partial w}{ \partial x} \frac{\partial x}{ \partial t} + \frac{\partial w}{ \partial y}\frac{\partial y}{ \partial t} = 2x \cdot (-s \sin t ) -2y \cdot ( s \cos t )$$
- $$ = - 2 \cdot 3 / \sqrt{2} \cdot 1 / \sqrt{2} - 2 \cdot 3 / \sqrt{2} \cdot 3 / \sqrt{2} = -18 $$
Exercise 13.5.20 Using Different Methods
Find \(\partial w / \partial s \) and \(\partial w / \partial t \) (a) by using the appropriate Chain Rule and (b) by converting w to a function for \(s\) and \(t\) before differentiating.
- \(w=x \cos yz, \: x=s^{2}, \: y=t^{2}, \: z=s-2t \)
Solution
- $$ \frac{\partial w}{ \partial s} = \frac{\partial w}{ \partial x} \frac{\partial x}{ \partial s} + \frac{\partial w}{ \partial y}\frac{\partial y}{ \partial s} + \frac{\partial w}{ \partial z}\frac{\partial z}{ \partial s} $$
- $$ = \cos (yz) \cdot 2s - xz \sin(yz) \cdot 0 - xy \sin(yz) \cdot 1 $$
- $$= 2s \cos(t^{2}(s-2t)) - s^{2}t^{2} \sin(t^{2}(s-2t)). $$
- $$ \frac{\partial w}{ \partial t} = \frac{\partial w}{ \partial x} \frac{\partial x}{ \partial t} + \frac{\partial w}{ \partial y}\frac{\partial y}{ \partial t} + \frac{\partial w}{ \partial z}\frac{\partial z}{ \partial t} $$
- $$ = \cos (yz) \cdot 0 - xz \sin(yz) \cdot 2t - xy \sin(yz) \cdot (-2) $$
- $$= -2s^{2}(s-2t)t \sin(t^{2}(s-2t)) + 2s^{2}t^{2} \sin(t^{2}(s-2t)). $$
Exercise 13.5.21 Finding a Derivative Implicitly
Differentiate implicitly to find \(dy / dx \).
- \( x^{2}-xy+y^{2}-x+y = 0 \)
Solution The identity defines \(y=y(x)\) implicitly. Differentiate both sides with respect to \(x\).
- $$ 2x-y-x \frac{dy}{dx} + 2y\frac{dy}{dx} -1 + \frac{dy}{dx} = 0 \Leftrightarrow (2x-y-1)-(x-2y-1) \frac{dy}{dx}=0 $$
Therefore,
- $$ \frac{dy}{dx} = \frac{2x-y-1}{x-2y-1} $$
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Parent Article: Calculus III Advanced (Course)
