#### n4rfhrd3r

##### New member
Hi,

I've been given this problem to work on as homework, and I'm completely confused by the wording.

"A stone is thrown vertically upwards and after t seconds its height y metres above its starting point is given by y=100t-5y^2. Calculate the greatest height reached by the stone and the value of t when this happens."

I need to solve it using calculus but I'm not sure how. I've differentiated the equation given to get dy/dx=100-10t, though I don't know if that's the right thing to do. Help appreciated

#### stapel

##### Super Moderator
Staff member
"A stone is thrown vertically upwards and after t seconds its height y metres above its starting point is given by y=100t-5y^2.
Is the last term really "5y^2"?

Calculate the greatest height reached by the stone and the value of t when this happens."
The question is just asking you to find the maximum value (the y-value) and the maximizing time (the t-value), given by the vertex. So you can check your work by using what you learned back in algebra.

I've differentiated the equation given to get dy/dx=100-10t
There is no "x" in the equation, so how are you differentiating with respect to x? Also, how did you get the derivative? Was the last term of the original function actually "5t^2"?

Thank you!

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#### HallsofIvy

##### Elite Member
To re-emphasize what staple just said: surely the height is $$\displaystyle y= 100t- 5t^2$$? You can find the maximum value by differentiating with respect to t and setting the derivative to 0 to find the time, t, when it is at the maximum height. His remark about "using what you learned back in algebra" refers to completing the square".

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#### n4rfhrd3r

##### New member
Is the last term really "5y^2"?

The question is just asking you to find the maximum value (the y-value) and the maximizing time (the t-value), given by the vertex. So you can check your work by using what you learned back in algebra.

There is no "x" in the equation, so how are you differentiating with respect to x? Also, how did you get the derivative? Was the last term of the original function actually "5t^2"?

Thank you!
Yes, the last term was 5t^2, I made a typo. Thanks for your help!

#### stapel

##### Super Moderator
Staff member
Yes, the last term was 5t^2.
Okay, so you have y(t) = 100t - 5t^2. Now differentiate with respect to t to get the appropriate expression for dy/dt (rather than "dy/dx").

Then: What did they give you as the process for finding the max/min points of a function, given the derivative of that function?