5. (5pts) find the slope ( y^{prime}(x) ) of the tangent line to the curve of ( y(x)=a x^{2}+b x+c ) where (…

5. (5pts) find the slope ( y^{prime}(x) ) of the tangent line to the curve of ( y(x)=a x^{2}+b x+c ) where ( a, b ), and ( c ) are constants. recall that the slope of the tangent line is given by ( y^{prime}(x)=\frac{y(x+delta x)-y(x)}{delta x} ) then let ( delta x \rightarrow 0 ) show all of your work. 6. (5pts) similarly, the velocity ( v(t) ) is the instantaneous rate of change (slope of the tangent line) of ( x(t) ) ( v(t)=\frac{x(t+delta t)-x(t)}{delta t} ) then let ( delta t \rightarrow 0 ) find the velocity ( v(t) ) given that ( x(t)=3 t^{2}+4 t ).

5. (5pts) find the slope ( y^{prime}(x) ) of the tangent line to the curve of ( y(x)=a x^{2}+b x+c ) where ( a, b ), and ( c ) are constants. recall that the slope of the tangent line is given by ( y^{prime}(x)=\frac{y(x+delta x)-y(x)}{delta x} ) then let ( delta x \rightarrow 0 ) show all of your work. 6. (5pts) similarly, the velocity ( v(t) ) is the instantaneous rate of change (slope of the tangent line) of ( x(t) ) ( v(t)=\frac{x(t+delta t)-x(t)}{delta t} ) then let ( delta t \rightarrow 0 ) find the velocity ( v(t) ) given that ( x(t)=3 t^{2}+4 t ).

Answer

Problem 5

Explanation:

Step1: Substitute (y(x)) into the formula

Given (y(x)=Ax^{2}+Bx + C), then (y(x+\Delta x)=A(x + \Delta x)^{2}+B(x+\Delta x)+C). Expand (y(x+\Delta x)): [ \begin{align*} y(x+\Delta x)&=A(x^{2}+2x\Delta x+\Delta x^{2})+Bx + B\Delta x+C\ &=Ax^{2}+2Ax\Delta x+A\Delta x^{2}+Bx + B\Delta x+C \end{align*} ] Now, (y(x+\Delta x)-y(x)=(Ax^{2}+2Ax\Delta x+A\Delta x^{2}+Bx + B\Delta x+C)-(Ax^{2}+Bx + C)) [ \begin{align*} y(x+\Delta x)-y(x)&=2Ax\Delta x+A\Delta x^{2}+B\Delta x\ \end{align*} ]

Step2: Calculate (\frac{y(x+\Delta x)-y(x)}{\Delta x})

[ \frac{y(x+\Delta x)-y(x)}{\Delta x}=\frac{2Ax\Delta x+A\Delta x^{2}+B\Delta x}{\Delta x}=2Ax + A\Delta x+B ]

Step3: Take the limit as (\Delta x\rightarrow0)

[ y^{\prime}(x)=\lim_{\Delta x\rightarrow0}\frac{y(x+\Delta x)-y(x)}{\Delta x}=\lim_{\Delta x\rightarrow0}(2Ax + A\Delta x+B) ] Since (\lim_{\Delta x\rightarrow0}(A\Delta x) = 0), we have (y^{\prime}(x)=2Ax + B)

Answer:

(y^{\prime}(x)=2Ax + B)

Problem 6

Explanation:

Step1: Substitute (x(t)) into the formula

Given (x(t)=3t^{2}+4t), then (x(t+\Delta t)=3(t + \Delta t)^{2}+4(t+\Delta t)) Expand (x(t+\Delta t)): [ \begin{align*} x(t+\Delta t)&=3(t^{2}+2t\Delta t+\Delta t^{2})+4t + 4\Delta t\ &=3t^{2}+6t\Delta t+3\Delta t^{2}+4t + 4\Delta t \end{align*} ] Now, (x(t+\Delta t)-x(t)=(3t^{2}+6t\Delta t+3\Delta t^{2}+4t + 4\Delta t)-(3t^{2}+4t)) [ \begin{align*} x(t+\Delta t)-x(t)&=6t\Delta t+3\Delta t^{2}+4\Delta t \end{align*} ]

Step2: Calculate (\frac{x(t+\Delta t)-x(t)}{\Delta t})

[ \frac{x(t+\Delta t)-x(t)}{\Delta t}=\frac{6t\Delta t+3\Delta t^{2}+4\Delta t}{\Delta t}=6t + 3\Delta t+4 ]

Step3: Take the limit as (\Delta t\rightarrow0)

[ v(t)=\lim_{\Delta t\rightarrow0}\frac{x(t+\Delta t)-x(t)}{\Delta t}=\lim_{\Delta t\rightarrow0}(6t + 3\Delta t+4) ] Since (\lim_{\Delta t\rightarrow0}(3\Delta t) = 0), we have (v(t)=6t + 4)

Answer:

(v(t)=6t + 4)