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This is one of our many quizzes on Slope fields. Click on this to access more such competitive quizzes on 2D Coordinate Geometry.

Slope fields Quiz 2

mathematics

 Kiran found a slope field for an unknown differential equation as shown below. He wants to find out whether the differential equation is a function of $y$ or function $x$ or the function of both $x,y$. What will be his answer? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5e52a524-6970-40f2-8166-8074ae41b3ed7825702015034050600.png"/>

 Check the given slope field for which the segments have the same slope either in row or column or both. If the row segments have the same slope then the differential equation is a function of $'x'$ and if column segments have the same slope then it is a function of $'y'$ and if both column and row segments have the same slope then it is a constant function.

Here, the segments of both row and column of the given slope field have the same slope. This means that the slope of the function throughout the graph is constant so that the differential equation is constant. $ \\ $ Therefore, the given differential equation is not a function of both $x,y$ and that is the differential equation is a constant function.

 A student found a differential equation in a book that is given as $\dfrac{dy}{dx}=x+y$. Now, he wants to draw the slope field of the differential equation that he found. What will be his required graph?

 Draw the graph of the given slope that is $x+y=0$ and check the options where the slope field has spreading curves in the direction of $x+y=0$ which will be the required answer.

The slope field will have curves in the direction of the graph of the given differential equation. $ \\ $ Given equation is $\dfrac{dy}{dx}=x+y$ $ \\ $ The graph of $x+y=0$ is given as, $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/eac56d13-1819-4f46-8fb0-f43c1e50dc8d8458345420322439036.png"/> $ \\ $ So, the slope field will have spreading curves in the direction of this equation. So, the slope field of the given differential equation is,$ \\ $  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e663a7c2-3973-4952-a873-d683dbd72960700249896400015703.png"/> $ \\ $ Only option $(a)$ matches the above slope field.

 Anush found a slope field for a specific differential equation as shown below and wants to know the differential equation that best represents the given slope field. What is the differential equation that represents the given slope field? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b9160f09-23ce-47bc-84dc-f36c2e4446e7603961384289717092.png"/>

 Check the graph whether the segments have the same slope for either row or column. If slopes are the same in row then it is a function of $'y'$ else it is a function of $'x'$ then check the sign of slopes for the respective interval and compare with the options for the correct differential equation.

From the graph the segments have same slope in a respected row across the graph so that the differential equation is the function of $'y'$ that is, $ \\ \dfrac{dy}{dx}=f\left( y \right) \\ $ From the graph the horizontal lines occur when $y=0,\pi ,-\pi $ but the function $f\left( y \right)={{y}^{2}}$ has horizontal lines only when $y=0$. So, option (a) is wrong. $ \\ $ The slopes in the graph are positive for $0  <  y  <  \pi $ and negative for $-\pi  < y < 0$ which corresponds to the graph of $f\left( y \right)=\sin y$. We know that this sign of slopes will be opposite for $f\left( y \right)=-\sin y$ $ \\ $ Therefore, the required differential equation is, $ \\ \dfrac{dy}{dx}=\sin y$

 The slope field of a respective differential equation having the solution $y=f\left( x \right)$ is shown below such that $f\left( 0 \right)=0$. Using the given slope field of the differential equation find the value of $f\left( 2 \right)$ $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d5a80ead-125e-428a-b71e-e718db78d3337953979005442386178.png"/>

 Take the initial value of the solution of the given differential equation and start the graph from that point following the segments from that point where we get the graph of solution of the differential equation. Then find the value on the $y-$axis for which the value on the $x-$axis is $2$ that gives the value of $f\left( 2 \right)$<sub>.</sub>

Given that the initial value of the function is $f\left( 0 \right)=0$ $ \\ $ So, the graph of $y=f\left( x \right)$ starts from the point $\left( 0,0 \right)$. Now let us draw the graph starts from $\left( 0,0 \right)$ and following the segments then we get the graph of $y=f\left( x \right)$ as, $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/f3886465-f456-49eb-8ff8-1571450bc36c692094247625618543.png"/> $ \\ $ The red line represents the graph of $y=f\left( x \right)$ such that when the value on $x-$axis is ‘2’, the value on $y-$axis is ‘2’ that is, $f\left( 2 \right)=2$

 The slope field of the differential equation having a solution as $y=f\left( x \right)$ is shown below. If is given that $f\left( 2 \right)=a$ when $f\left( 0 \right)=0$ and $f\left( 1 \right)=b$ when $f\left( 0 \right)=1$. Find the quadratic equation that is having the roots as $a,b$. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/73e21d77-d2fa-43d0-92b9-4ac7354f3dca7133785361494213343.png"/>

 Draw the graph of $y=f\left( x \right)$ from the initial value by following the segments and then find the value of $a,b$ from the graph. Then the quadratic equation having roots $\alpha ,\beta $ is given as ${{x}^{2}}-\left( \alpha +\beta  \right)x+\alpha \beta =0$ to get the required answer.

First find the value of $'a'$ $ \\ $ The initial value is given as $f\left( 0 \right)=0$. So, the graph of $y=f\left( x \right)$ starts from the point $\left( 0,0 \right)$. Now let us draw the graph starts from $\left( 0,0 \right)$ and following the segments then we get the graph of $y=f\left( x \right)$ as, $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4e8fb5c5-46b7-4d32-85ea-b8d14363316d374345029118071197.png"/> $ \\ $ The red line represents the graph of $y=f\left( x \right)$ when $f\left( 0 \right)=0$ such that when the value on $x-$axis is ‘2’, the value on $y-$axis is ‘4’ that is, $f\left( 2 \right)=4=a$<sub> </sub>$ \\ $ Now, the value of $'b'$ $ \\ $ The initial value is given as $f\left( 0 \right)=1$. So, the graph of $y=f\left( x \right)$ starts from the point $\left( 0,1 \right)$. Now let us draw the graph starts from $\left( 0,1 \right)$ and following the segments then we get the graph of $y=f\left( x \right)$ as, $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7e8e8f3b-400b-471f-9e8e-7b405f2bc05e9034717222443347411.png"/> $ \\ $ The red line represents the graph of $y=f\left( x \right)$ when $f\left( 0 \right)=1$ such that when the value on $x-$axis is ‘1’, the value on $y-$axis is ‘2’ that is, $f\left( 1 \right)=2=b$ $ \\ $ Now, the quadratic equation having roots $a,b$ is, $ \\ \begin{align}  & \Rightarrow {{x}^{2}}-\left( a+b \right)x+ab=0 \\  & \Rightarrow {{x}^{2}}-\left( 4+2 \right)x+\left( 4\times 2 \right)=0 \\  & \Rightarrow {{x}^{2}}-6x+8=0 \\ \end{align}$